ANALYTIC  GEOMETRY  AND 
CALCULUS 


BY 
FREDERICK   S.  WOODS 

ANI» 

FREDERICK   II.  BAILEY 


I'Kol ■  |.;.>m.i:s    ol"    MATllKMVI'lis    IN    TIIK 
MASSACHUSETTS    l\-llli   ll.   01     I  I  I  M  Noi.ucv 


GINN  AND  COMPANY 

BOSTON     •    NEW    YORK     •     CHICAGO     •     LONDON 
ATLANTA     •     DALLAS     •     COLUMBUS     •     SAN    FRANCISCO 


COPYRIGHT,  1017,  BY  FREDERICK  S.  WOODS 
AND  FREDERICK  H.  BAILEY 

ALL   KIUHTS  RESERVED 
217.2 


GINN  AND  COMPANY  •  PRO- 
PRIETORS •  BOSTON  •  U.S.A. 


PREFACE 


The  present  work  is  a  revision  and  abridgment  of  the  authors' 
"Course  in  Mathematics  for  Students  of  Engineering  and  Applied 
Science."  The  condensation  of  a  two-volume  work  into  a  single 
volume  has  been  made  possible  partly  by  the  omission  of  some 
topics,  but  more  especially  by  a  rearrangement  of  subject  matter 
and  new  methods  of  treatment. 

Among  the  subjects  omitted  are  determinants,  much  of  the 
general  theory  of  equations,  the  general  equation  of  the  conic 
sections,  polars  and  diameters  related  to  conies,  center  of  curv- 
ature, evolutes,  certain  special  methods  of  integration,  complex 
numbers,  and  some  types  of  differential  equations.  All  these 
subjects,  while  interesting  and  important,  can  well  be  postponed 
to  a  later  course,  especially  as  their  inclusion  in  the  present 
course  would  mean  the  crowding  out,  or  less  thorough  han- 
dling, of  subjects  which  are  more  immediately  important. 

The  rearrangement  of  material  is  seen  especially  in  the  bring- 
ing together  into  the  first  part  of  the  book  of  all  methods  for 
the  graphical  representation  of  functions  of  one  variable,  both 
algebraic  and  transcendental.  This  has  the  effect  of  devoting 
the  first  part  of  the  book  to  analytic  geometry  of  two  dimen- 
sions, the  analytic  geometry  of  three  dimensions  being  treated 
later  when  it  is  required  for  the  study  of  functions  of  two 
variables.  The  transition  to  the  calculus  is  made  early  through 
the  discussion  of  slope  and  area  (Chapter  IX  ),  the  student  being 
thus  introduced  in  the  first  year  of  his  course  to  the  concepts 
of  a  derivative  and  a  definite  integral  as  the  limit  of  a  sum. 

The  new  methods  of  handling  the  subject  matter  will  be 
recognized  by  the  teacher  in  places  too  numerous  to  specify 
here.  The  articles  on  empirical  equations,  the  remainder  in 
Taylor's  series,  and  approximate  integration  are  new. 


zT^nm  /*. « 


iv  PKEFACE 

It  is  believed  that  this  book  can  be  completely  studied  by 
an  average  college  class  in  a  two  years'  course  of  180  exer- 
cises. Teachers  who  wish  a  slower  pace,  however,  may  omit 
the  last  chapter  on  differential  equations,  or  substitute  it  for 
some  of  the  work  on  multiple  integrals. 

The  book  contains  2000  problems  for  the  student,  many  of 
which  are  new.  It  is,  of  course,  not  expected  that  any  student 
will  solve  all  of  them,  but  the  supply  is  ample  enough  to  allow  the 
assignment  of  different  problems  for  home  work  and  classroom 
exercises,  and  to  allow  different  assignments  in  successive  years. 

F.  S.  WOODS 
F.  H.  BAILEY 


COXTEXTS 


Chapter  I  — CARTESIAN  COORDINATES 
Article  Page 

1.  Direction  of  a  straight  line 1 

2.  Projection 2 

3.  Number  scale 3 

4.  Coordinate  axes 4 

5.  Distance 5 

6.  Slope G 

7.  Point  of  division 8 

S.  Variable  and  function 9 

9.  Functional  notation 12 

Problems 18 


Chapter  II— GRAPHS  <>F  ALGEBRAIC  FUNCTIONS 

10.  Equation  and  graph 20 

11.  Intercepts 21 

12.  Symmetry  and  impossible  values 28 

13.  Infinite  values 25 

11.   Intersection  of  graphs 29 

15.  Real  roots  of  an  equation 3  I 

Problems 36 


Chapter  III  — CHANGE  OF  COORDINATE  AXES 

16.  Introduction 39 

17-18.  Change  of  origin 39 

19.  Change  of  direction  of  axes 42 

20.  Oblique  coordinates 43 

21.  Change  from  rectangular  to  oblique  axes 44 

22.  Degree  of  the  transformed  equation 45 

Problems 45 


vi  CONTENTS 

Chapter  IV  — GRAPHS  OF  TRANSCENDENTAL  FUNCTIONS 
Article  Page 

23.  Definition 49 

24.  Trigonometric  functions 49 

25.  Inverse  trigonometric  functions 52 

26.  Exponential  and  logarithmic  functions 52 

27.  The  number  e 53 

Problems ' 55 

Chapter  V  — THE  STRAIGHT  LINE 

28.  The  pointrslope  equation 57 

29.  The  slope-intercept  equation 57 

30.  The  two-point  equation 58 

31.  The  general  equation  of  the  first  degree 58 

32.  Angles 59 

33.  Distance  of  a  point  from  a  straight  line 63 

Problems 64 

Chapter  VI  — CERTAIN  CURVES 

34.  Locus  problems 69 

35-37.  The  circle 70 

38-40.  The  ellipse 74 

41-43.  The  hyperbola 77 

44-45.  The  parabola 80 

46.  The  conic 83 

47.  The  witch 84 

48.  The  cissoid 85 

49.  The  strophoid 86 

50.  Use  of  the  equation  of  a  curve 88 

51.  Empirical  equations 89 

Problems 92 

Chapter  VII —  PARAMETRIC  REPRESENTATION 

52.  Definition 106 

53.  The  circle 108 

54.  The  ellipse 108 

55.  The  cycloid 109 

56.  The  trochoid 110 


CONTENTS  vii 

Article  Page 

57.  The  epicycloid Ill 

58.  The  hypocycloid 112 

59.  The  involute  of  the  circle 112 

Problems 113 

Chapter  VIII  — POLAR  COORDINATES 

60.  Coordinate  system 118 

61.  The  spirals 120 

62.  The  straight  line 121 

63.  The  circle 122 

64.  The  limacon 123 

65.  Relation  between  rectangidar  and  polar  coordinates      .     .     .  124 

66.  The  conic,  the  focus  being  the  pole 125 

67.  Examples 126 

Problems 127 

Chapter  IX— SLOPES  AND  AREAS 

68.  Limits 130 

69.  Theorems  on  limits L32 

70.  Slope  of  a  curve 134 

71.  Increment 135 

72.  Continuity L36 

7:!.  Derivative L36 

71.   Differentiation  of  a  polynomial L37 

75.  Sign  of  tin'  derivative L38 

76.  Tangent  line 140 

77.  The  differential 141 

78.  Area  under  a  curve 1  13 

79.  Differential  of  area 1  l<i 

80.  The  integral  of  a  polynomial 146 

81:  The  definite  integral 147 

Problems 150 

Chapter  X  — DIFFERENTIATION  OF  ALGEBRAIC   FUNCTIONS 

82.  Theorems  on  derivatives 151 

83.  Derivative  of  u" 159 

84.  Formulas 161 

85.  Higher  derivatives 162 

86.  Differentiation  of  implicit  functions 162 


vm 


CONTENTS 


Article                                                                               •  Page 

87.  Tangents  and  normals 164 

88.  Sign  of  the  second  derivative 166 

89.  Maxima  and  minima 168 

90.  Limit  of  ratio  of  arc  to  chord 172 

91.  The  differentials  dx,  <///,  ds 174 

92.  Rate  of  change 175 

it:').   Rectilinear  motion 177 

94.  Motion  in  a  curve 180 

Problems 181 

Chapter  XI  — DIFFERENTIATION  OF  TRANSCENDENTAL 
FUNCTIONS 

95.  Limit  of  ^^ 192 

h 

96.  Differentiation  of  trigonometric  functions 193 

97.  Differentiation  of  inverse  trigonometric  functions      .      .     .  196 

l 

98.  Limit  of  (1  +  ft)* 199 

99-100.   Differentiation  of  exponential  and  logarithmic  functions    .  199 

101.  Applications 202 

102.  The  derivatives  in  parametric  representation 204 

10:5.  Direction  of  a  curve  in  polar  coordinates 205 

104.  Derivatives  with  respect  to  the  arc  in  polar  coordinates      .  206 

105.  Curvature 207 

106.  Radius  of  curvature 208 

107.  Radius  of  curvature  in  parametric  representation       .      .      .  210 

108.  Radius  of  curvature  in  polar  coordinates 211 

Problems 212 

Chapter  XII  — INTEGRATION 

109.  Introduction 222 

110.  Fundamental  formulas 222 

111.  Integral  of  u» 223 

112.  Integrals  of  trigonometric  functions 226 

113-114.    Integrals  leading  to  inverse  trigonometric  functions       .      .  2:50 

115.    Integrals  of  exponential  functions 236 

110.   Collected  formulas 236 

117-118.    Integration  by  substitution 238 

119-120.   Integration  by  parts 243 

121.  Integration  by  partial  fractions 247 

122.  Reduction  formulas 252 

Problems 253 


CONTENTS  ix 

Chapter  XIII  — APPLICATIONS  OF  INTEGRATION 

Article  Page 

123-124.  Element  of  a  definite  integral 260 

125.  Area  of  a  plane  curve  in  Cartesian  coordinates     ....  262 

126.  Infinite  limits  or  integrand         261 

127.  The  mean  value  of  a  function 265 

12s.  Area  of  a  plane  curve  in  polar  coordinates 267 

129.  Volume  of  a  solid  with  parallel  liases 268 

130.  Volume  of  a  solid  of  revolution 270 

131-1:52.  Length  of  a  plane  curve 272 

133.  Area  of  a  surface  of  revolution '271 

131.  Work 275 

135.  Pressure 275 

136.  Center  of  pressure 277 

137.  Center  of  gravity 278 

138.  Attraction 283 

Problems 285 

Chafteb   XIV      SPACE  GEOMETRY 

130.  Functions  of  more  than  one  variable 300 

110.  Rectangular  coordinates 301 

141.  Graphical  representation  of  a  function  of  two  variables  30] 

142.  Cylinders 303 

143.  Other  surfaces 304 

111.  Surfaces  of  revolution 309 

145.  Projection 310 

146.  Components  of  a  directed  straight  line 312 

117.  Distance  between  two  points 313 

148.  Direction  cosines 31  1 

149.  Angle  between  two  straight  lines 315 

150.  Direction  of  the  normal  to  a  plane 316 

151.  Equation  of  a  plane  through  a  given  point  perpendicular  to 

a  given  direction 316 

152.  Angle  between  two  planes 316 

153.  Equations  of  a  straight  line -">17 

154.  Straight  line  determined  by  two  points 317 

155.  Straight  line  passing  through   a  known  point  in  a  given 

direction 318 

156.  Determination  of  the  direction  cosines  of  a  straighl  line    .  319 

157.  Distance  of  a  point  from  a  plane 320 

158.  Problems  on  the  plane  and  the  straight  line 321 


x  CONTENTS 

Article  Page 

159.  Space  curves 322 

160.  Direction  of  space  curve  and  element  of  arc 324 

161.  Tangent  lint'  and  normal  plane 326 

Problems 32G 

Chapter  XV  — PARTIAL  DIFFERENTIATION 

162.  Partial  derivatives 335 

163.  Higher  partial  derivatives 338 

164.  Increment  and  differential  of  a  function  of  two  variables  .  339 

165.  Extension  to  three  or  more  variables 342 

166.  Directional  derivative  of  a  function  of  two  variables      .     .  343 

167.  Total  derivative  of  z  with  respect  to  a; 344 

168.  The  tangent  plane 345 

169.  Maxima  and  minima 348 

170.  Exact  differentials 349 

171.  Line  integrals 353 

172.  Differentiation  of  composite  functions .  357 

Problems ' 361 

Chapter  XVI  — MULTIPLE  INTEGRALS 

173.  Double  integral  with  constant  limits 369 

174.  Double  integral  with  variable  limits 371 

175.  Computation  of  a  double  integral 373 

176.  Double  integral  in  polar  coordinates 374 

177.  Area  bounded  by  a  plane  curve 376 

178.  Moment  of  inertia  of  a  plane  area 377 

179.  Center  of  gravity  of  plane  areas 379 

180.  Area  of  any  surface 381 

181.  Triple  integrals 385 

182.  Change  of  coordinates 388 

183.  Volume 389 

184.  Moment  of  inertia  of  a  solid 390 

185.  Center  of  gravity  of  a  solid 392 

186.  Attraction 393 

Problems 394 

Chapter  XVII  —  INFINITE  SERIES 

187.  Convergence 405 

188.  The  comparison  test  for  convergence 406 

189.  The  ratio  test  for  convergence 407 


CONTEXTS  xi 

Article  Page 

190.  Absolute  convergence 409 

191.  The  power  series 410 

192.  Maclaurin's  and  Taylor's  series 412 

193.  The  remainder  in  Taylor's  series 416 

194.  Relations  between  the  exponential  and  the  trigonometric 

functions 418 

195.  Approximate  integration 419 

196.  The  theorem  of  the  mean 422 

197.  The  indeterminate  form  J} 123 

198.  Other  indeterminate  forms \-j:> 

199.  Fourier's  series 127 

Problems 1.">1 

Chapteb  XVIII  —  DIFFERENTIAL   EQUATIONS 

200.  Definitions |:',s 

201.  The  equation  Mdx  +  Ndy  =  0,  when  the  variables  can  be 

separated Ill 

202-203.  The  homogeneous  equation  Mdx  +  Ndy=0 142 

204.  The  linear  equation  of  the  first  order      .     .          ....  Ill 

205.  Bernoulli's, equation llii 

206.  The  exact  equation  Mdx  +  Ndy  =  0 117 

207.  The  integrating  factor lis 

'Jos.  Certain  equations  of  the  second  order 149 

209.  The  linear  equation  with  constant  coefficients 154 

210.  The    linear    equation    of   tin'    first    order    with    constant 

coefficients        \~>~< 

211.  The  linear  equation   of  the  second  order  with   constant 

coefficients t56 

212.  The  general  linear  equation  with  constant  coefficients   .  160 
21:5.  Solution  by  undetermined  coefficients L62 

214.  Systems    of    linear    differential    equations    with    constant 

coefficients 165 

215.  Solution  by  series 166 

Problems 471 

Answers 481 

Index  .514 


ANALYTIC  GEOMETRY  AND 
CALCULUS 

CHAPTER  I 

CARTESIAN  COORDINATES 

1.  Direction  of  a  straight  line.  Consider  any  straight-  line 
connecting  two  points  A  and  B.  In  elementary  geometry,  only 
the  position  and  the  length  of  the  line  are  considered,  and 
consequently  it  is  immaterial  whether  the  line  be  called  AB  or 
BA;  but  in  the  work  to  follow,  it  is  often  important  to  consider 
the  direction  of  the  line  as  well.    Accordingly,  if  the  direction 

of  the  line  is  considered  as  from  , , , 

A  to  B  it  is  called  AB,   but  if  A  BO 

Fn;     1 

the   direction   is  considered  from 

B  to  A  it  is  called  BA.  It  will  be  seen  later  that  the  distinc- 
tion between  A3  and  HA  is  the  same  as  that  between  -\-a 
and  —  a  in  algel  >ra. 

Consider  now  two  segments  AB  and  BC  on  the  same  straight 
line,  the  point  B  being  the  end  of  the  first  segment  and  the 
beginning  of  the  second.  The  segment  AC  is  called  the  sum  of 
AB  and  BC  and  is  expressed  by  the  equation 

AB  +  BC=AC.  (1) 

This  is  clearly  true  if  the  points  are  in  the  position  of  fig.  1,  but 
it  is  equally  true  when  the  points  are  in  the  position  of  fig.  2. 

Here  the  line  BC,  being  opposite 

in  direction  to  AB,  cancels  part  of         a  c  b 

it,  leaving  AC.  FlG-  2 

If,  in  the  last  figure,  the  point  C  is  moved  toward  A,  the 
sum  AC  becomes  smaller,  until  finally,  when  C  coincides  with  A, 
we  have  jb  +  BA  =  0,     or     BA  =  -AB.  (2) 

l 


2  CARTESIAN  COORDINATES 

If  the  point  C  is  at  the  left  of  A,  as  in  fig.  3,  we  still  have 
AB  +  BC  =  AC,  where  AC=~  CA  by  (2). 

It  is  evident  that  this  addition  is  analogous  to  algebraic 
addition,  and  that  this  sum  may  be  an  arithmetical  difference. 

From  (1)  we  may  obtain  by 
transposition  a  formula  for  sub-  C  A  b 

traction  ;  namely,  Fig.  3 

BC  =  AC-AB.  (3) 

This  is  universally  true,  since  (I)  is  universally  true. 

2.  Projection.  Let  AB  and  MN  (figs.  4,  5)  be  any  two  straight 
lines  in  the  same  plane,  the  positive  directions  of  which  are 
respectively  AB  and  MN.    From  A  and  B  draw  straight  lines 


perpendicular  to  MN,  intersecting  it  at  points  A'  and  B'  respec- 
tively. Then  A'B1  is  the  projection  of  AB  on  MN,  and  is  positive 
if  it  has  the  direction  MN  (fig.  4),  and  negative  if  it  has  the 
direction  NM  (fig.  5). 

Denote  the  angle  between  MN  and  AB  by  </>,  and  draw  AC 
parallel  to  MN.    Then  in  both  cases,  by  trigonometry, 

AC  =  AB  cos <f>. 

But  AC  =  A'B',  and  therefore 

A'B' =AB  cos  (f>. 

Hence,  to  find  the  projection  of  one  straight  line  upon  a  second, 
multijrty  the  length  of  'the  first  by  the  cosine  of  the  angle  between 
the  positive  directions  of  the  two  lines. 

The  projection  of  a  broken  line  upon  a  straight  line  is  defined 
as  the  algebraic  sum  of  the  projections  of  its  segments. 


NUMBER  SCALE  3 

Let  ABODE  (fig.  6)  be  a  broken  line,  MX  a  straight  line 
in  the  same  plane,  and  AE  the  straight  line  joining  the  ends 
of  the  broken  line. 

Draw  AA',  BB',  CO',  DD',  and 
EE'  perpendicular  to  MN;  then 
A'B',  B'C,  CD',  D'E',  tm&A'E' 
are  the  respective  projections  on 
JAY  of  AB,  BC,  CD,  DE,  and  JA. 

But,  by  §  1,  A'B'  +  B'  C'  +  (  "D1  +  D'E'  =  A '  /•/'. 

Hence  the  projection  of  a  broken  line  upon  a  straight  line  is 
equal  to  the  projection  of  the  straight  line  joining  the  extremities 
of  the  broken  line. 

3.  Number  scale.  On  any  straight  line  assume  a  fixed  point 
0  as  the  zero  point,  or  origin,  and  lay  off  positive  numbers  in 
one  direction  and  negative  numbers  in  the  other.  If  the  line 
is  horizontal,  as  in  fig.  7,  it  is  usual,  hut  not  necessary,  to  lay 
off  the  positive  numbers  to  the  right  of  0  and  the  negative 
numbers  to  the  left.  The  numbers  which  we  can  thus  lay  off 
are  of  two  kinds:  the  rational  num-  „ 

hers,  including  the  integers  and  the  _I  _ '.  _\  _\ — J — \ — £ — ^-^ — 
common  fractions;  and  the  irrational  F  ,    _ 

numbers,  which  cannot  be  expressed 

exactly  as  integers  or  common  fractions,  but  which  may  be  so 
expressed  approximately  to  any  required  degree  of  accuracy. 
The  rational  and  the  irrational  numbers  together  form  the  class 
of  real  numbers. 

Then  any  point  M  on  the  scale  represents  a  real  number, 
namely,  the  number  which  measures  the  distance  of  M  from  0: 
positive  if  M  is  to  the  right  of  0,  and  negative  if  M  is  to  the 
left  of  0.  Conversely,  any  real  number  is  represented  by  one 
and  only  one  real  point  on  the  stale. 

Imaginary,  or  complex,  numbers,  which  are  of  the  form  a+hs/—  1, 
cannot  be  laid  off  on  the  number  scale. 

The  result  of  §  1  is  particularly  important  when  applied  to 
segments  of  the  number  scale.  For  if  x  is  any  number  corre- 
sponding to  the  point  M,  we  may  always  place  x  =  031,  since  both 


CARTESIAN  COORDINATES 


x  and  OM  are  positive  when  M  is  at  the  right  of  0,  and  both 
x  and  OM  are  negative  when  M  is  at  the  left  of  0.    Now  let  M 
and  .!/.,  (fig.  8)  be  any  two  points,  and  let  xx=  OM  and  %  —  OM. 
Then 

o     Mx  M„ 


OM  =  x2 

On  the  other  hand, 

MM  =  OJf 


-10     12    3 

Fig.  8 


03/ 


It  is  clear  that  the  segment  MXM2  is  positive  when  M2  is  at 
the  right  of  Mx,  and  negative  when  i!f2  is  at  the  left  of  Mx. 

Hence  the  length  and  the  sign  of  any  segment  of  the  number 
scale  is  found  by  subtracting  the  value  of  the  x  corresponding  to 
the  beginning  of  the  segment  from  the  value  of  the  x  corresponding 
to  the  end  of  the  segment. 

4.  Coordinate  axes.  Let  OX  and  OY  be  two  number  scales  at 
right  angles  to  each  other,  with  their  zero  points  coincident  at  0, 
as  in  fig.  9.  y 

Let  P  be  any  point  in  the  plane,  - 
and  through  P  draw  straight  lines 
perpendicular  to  OX  and  OY  re- 
spectively, intersecting  them  at  M  . 
and  N.  If  now,  as  in  §  3,  we 
place  x  =  OM  and  y  =  OX,  it  is 
clear  that  to  any  point  P  there 
corresponds  one  and  only  one 
pair  of  numbers  x  and  y,  and 
to  any  pair  of  numbers  corresponds  one  and  only  one  point  P. 

If  a  point  P  is  given,  x  and  y  may  be  found  by  drawing  the 
two  perpendiculars  MP  and  XP,  as  above,  or  by  drawing  only 
one  perpendicular  as  MP.    Then 

0M=x,         MP=0X=y. 

On  the  other  hand,  if  x  and  y  are  given,  the  point  P  may  be. 
located  by  finding  the  points  M  and  X  corresponding  to  the  num- 
bers x  and  y  on  the  two  number  scales  and  drawing  perpendiculars 
to  OX  and  0  Y  respectively  through  M  and  X.  These  perpen- 
diculars intersect  at  the  required  point  P.    Or,  as  is  often  more 


Fig.  9 


DISTANCE 


convenient,  a  point  21  corresponding  to  x  may  be  located  on  its 
number  scale,  and  a  perpendicular  to  OX  may  be  drawn  through 
31,  and  on  tins  perpendicular  the  value  of  y  laid  off.  In  fig.  9, 
for  example,  31  (corresponding  to  x)  may  be  found  on  the  scale 
OX,  and  on  the  perpendicular  to  OX  at  M,  MI'  may  lie  laid  off 
equal  to  y.  When  the  point  is  located  in  either  of  these  ways  it  is 
said  to  be  plotted.  It  is  evident  that  plotting  is  most  conveniently 
performed  when  the  paper  is  ruled  in  squares,  as  in  fig.  9. 

These  numbers  ./■  and  //  are  called  respectively  the  abscissa  and 
the  ordinate  of  the  point,  and  together  they  are  called  its  coordi- 
nates. It  is  to  be  noted  that  the  abscissa  and  the  ordinate,  as 
defined,  are  respectively  equal  to  the  distances  from  <>  V  and  OX 
to  the  point,  the  direction  as  well  as  the  magnitude  of  the  distances 
being  taken  into  account.  Instead  of  designating  a  point  by  writing 
x  =  a  and  y  =  —  b,  it  is  customary  to  write  /'( <t,  —  l>),  the  abscissa 
always  being  written  first  in  the  parenthesis  and  separated  from 
the  ordinate  by  a  comma.  OX  and  OY  are  called  the  axes  of 
coordinates,  but  are  often  referred  to  as  the  axes  of  x  and  y 
respectively. 

5.  Distance.  Let  I^0\,  y^  and  /_!(./„,  //..)  be  two  points,  and 
at  first  assume  that  /;/.'  is  parallel  to  <>ne  of  the  coordinate  axes, 
as  <>X  (tig.  Hi).  Then  //,  =  //,.  Now  .!/, .)/.,, 
the  projection  of  /',/'.,  on  OX,  is  evidently 
equal    to    %Py      15ut  "j/J/,=  .r,- ^   (§  3). 


nence                  P^x-x^ 

(1) 

In  like  manner,  if  ./;,=  x ,  I\P3 

is  parallel 

to  OF,  and         ^==yg_yi. 

(2) 

p, 

Pn 

; 

Jtf,     0 

M,    ' 

Fig.  10 


If  x2=f=  xx  and  //o^  yx,  I\r,  is  not  parallel  to  either  axis,  bet 
the  points  be  situated  as  in  tig.  11,  and 
through  ]'.  and  /.!  draw  straight  lines 
parallel  respectively  to  OX  and  0  )'.  They 
will  meet  at  a  point  7/,  the  coordinates 
of  which  are  readily  seen  to  be  (./.,,  y^. 
By  (1)  and  (2), 


J[U  =  x  —  x , 


in: 


.",• 


CARTESIAN  COORDINATES 
But  in  the  right  triangle  PlRP2, 


whence,  by  substitution,  we  have 


tt-V(*t-*Oa+(yt-tf.  (3) 

It  is  to  be  noted  that  there  is  an  ambiguity  of  algebraic  sign, 
on  account  of  the  radical  sign.  But  since  PXP2  is  parallel  to 
neither  coordinate  axis,  the  only  two  directions  in  the  plane  the 
positive  directions  of  which  have  been  chosen,  we  are  at  liberty 
to  choose  either  direction  of  I[P2  as  the  positive  direction,  the 
other  becoming  the  negative. 

It  is  also  to  be  noted  that  formulas  (1)  and  (2)  are  particular 
cases  of  the  more  general  formula  (3). 

Ex.  Find  the  coordinates  of  a  point  equally  distant  from  the  three  points 
P1(l,2),P^-i;-2),«ndPi(2,-5). 

Let  P(x,  y)  be  the  required  point.    Then 

PXP  =  I\_P    and    P„P  =  P3P. 


But  PXP  =  V(x  -iy2  +  (y-  2)2, 


P2P  =  V(x  +  l)2  +  (y  +  2)s 


PSP  =  V(x  -  2f  +  (//  +  5)2. 


Therefore       V(z  -  l)2  +  (y  -  2)2  =  V(x  +  l)'2  +  (y  +  2)2, 
V(x  +  l)2  +  (//  +  2)2  =  V(ar  -  2)2  +  (y  +  5)2 ; 

whence,  by  solution,  x  =  |  and  y  =  —  f .    Therefore  the  required  point  is 
(I,  -  I)- 

6.  Slope.  Let  %(xx,  y^  and  P2(x2,  y2)  (figs.  12,  13)  be  two 
points  upon  a  straight  line.  If  we  imagine  that  a  point  moves 
along  the  line  from  P^  to  P2,  the  change  in  x  caused  by  this 
motion  is  measured  in  magnitude  and  sign  by  x2  —  xv  and  the 
change  in  y  is  measured  by  y%—  yx.  We  define  the  slope  of  the 
straight  line  as  the  ratio  of  the  change  in  y  to  the  change  in  x  as 
a  point  moves  along  the  line,  and  shall  denote  it  by  the  letter  m. 
We  have  then,  by  definition, 

V»  —  Vi 
m  =  ^—^l.  (1) 


SLOPE 


A  geometric  interpretation  of  the  slope  is  readily  given.  For 
if  we  draw  through  1[  a  line  parallel  to  OX,  and  tlirough  P2  a 
line  parallel  to  OY,  and  call  R  the  point  in  which  these  two 
lines  intersect,  then  x2  —  xx  =  J^E  and  y2  —  yl  =  RP2 ;  and  hence 

It  is  clear  from  the  figures  that  the  value  of  m  is  independent 
of  the  two  points  J[  and  I2  and  depends  only  on  the  given  line. 
We  may  therefore  choose  1[  and  J',  (as  in  figs.  12  and  13)  so 
that  1{R  is  positive.     There  are  then  two  essentially  different 


Fig.  12 


Fig.  13 


cases,  according  as  the  line  runs  up  or  down  toward  the  right 
hand.  In  the  former  case;  ///_;  and  m  are  positive  (tig.  12);  in 
the  latter  case  1U\  and  m  are  negative  (fig.  13).  We  may  state 
this  as  follows: 

The  slope  of  a  straight  line  is  positive  when  an  increase  in  x 
causes  an  increase  in  y,  and  is  negative  when  an  turn-use  in  x 
causes  a  decrease  in  y. 

When  the  line  is  parallel  to  OX,  y  =  y ,  and  consequently  m  =  0. 
If  the  line  is  parallel  to  0 )',  ./•.,=  x^,  and  therefore  m  =  oo  (§  13). 

Ex.    Find  a  point  distant  5  units  from  the  point  (1,  —2)  and  situated 
so  that  the  slope  of  the  straight  line  joining  it  to  (1,  —  2)  is  g. 
Let  P(x,  y)  be  the  required  point.    Then 

(x  -  l)2  +  0/  +  2)2  =  25, 

y±l  =  i 

x-l      3 

Solving  these  two  equations,  we  find  two  points,  (i,  2)  and  (—  2,  —  6). 


8 


CART  ES  LAN  ( !0<  >  RD1NATES 


7.  Point  of  division.    Let  P(x,  if)  be  a  point  on  the  straight 
line  determined  by  ll(*\,  y^)  an(l  P(.X2>  #»)»  s0  situated  that 

There  are  three  cases  to  consider,  according  to  the  position  of 
the  point  P.    If  P  is  between  the  points  7/  and  J£(fig.  14),  the 


Jf,   if    Jfg 


Fig.  15 


segments  i^P  and  P1P2  have  the  same  direction,  and  PfP  <  7^ ; 
accordingly  Z  is  a  positive  number  less  than  unity.  If  P  is 
beyond  P2  from  if  (fig.  15),  PXP  and  PXP2  still  have  the  same 
direction,  but  P1P  >  PXP,  ;  therefore  /  is  a  positive  number 
greater  than  unity.  Finally,  if  P  is  beyond  P1  from  B,  (fig.  16), 
PXP  and  J[P2  have  opposite  directions, 
and  I  is  a  negative  number,  its  nu- 
merical value  ranging  all  the  way  from 
0  to  00. 

In  the  first  case  P  is  called  a  point 
of  internal  division,  and  in  the  last 
two  cases  it  is  called  a  point  of  ex- 
ternal division. 

In  all  three  figures  draw  PXMV  PM, 
and  P2M2  perpendicular  to  OX.    In  each  figure  0M=  OMx  +  MxM\ 
and  since  PXP  =l(PxP^),  MvM=l(MiMtf)  by  geometry. 

Therefore  0M=  0M1  +  I  (^  JQ ; 

whence,  by  substitution, 

x  —  xy  +  I  (x2  —  xf).  (1) 

By  drawing  lines  perpendicular  to  07we  can  prove,  in  the 
same  way,  y  =  y  +  /  („  _  „  ).  (2) 


M  o 


M,    M, 


Fig.  16 


VARIABLE  AXD  FUNCTION 


9 


In    particular,    if 
formulas  become 

x 


bisects    the   line   7//',    /  =  },   and  these 
V\  +  .'i-i 


•ri  +  X-2 

9. 


II 


Ex.  1.    Find  the  coordinates  of  a  point  §  of  the  distance  from  J\(2,  3) 
to  ]'.,(•■),  -:!). 

If  the  required  point  is  P(x,  y), 

x  =  2  +  |(3 -2)  =  2§, 
y  =  3  +  f(-3-3)=|. 

Ex.  2.    Prove   analytically   that   the    straight   line    dividing    two    sides 
of  a  triangle  in  the   same   ratio   is  parallel   to  the   third  side. 

Let  one  side  of  the  triangle  coincide 
with  OX,  one  vertex  being  at  0.  Then  tin; 
vertices  of  the  triangle  air  0(0,  0),  A(xv  ••), 
B(x2,  //,)  (fig.  17).  Let  CD  divide  the  aides 
OB  and  AB  so  that  0C  =  l(OB)  and 
AD  =  I  {All). 

If  the  coordinates  of  C  arc  denoted  by 
(z8,  //..)  and  those  of  1>  by  (..-.,,  //.,),  then,  by 
the  above  formulas, 

x3  =  lxv 

xi  =  x1  +  /(■':,  -xx) 

Since  ?/.,  =  i/i,  CI)  is  parallel  to  0A. 


X 


Fig.  17 


8.  Variable   and   function.    A    quantity    which    remains    un- 
changed throughout  a  given  problem  or  discussion  is  called  a 

constant.  A  quantity  which  changes  its  value  in  the  course  of 
a  problem  or  discussion  is  called  a  variable.  If  two  quantities 
are  so  related  that,  when  the  value  of  one  is  given  the  value  of 
the  other  is  determined,  the  second  quantity  is  called  a  function 
of  the  first.  When  the  two  quantities  are  variables,  the  first  is 
called  the  independent  variable^  and  the  function  is  sometimes 
called  the  dependent  variable.  As  a  matter  of  fact,  when  two 
related  quantities  occur  in  a  problem,  it  is  usually  a  matter  of 
choice  which  is  called  the  independent  variable  and  which  the 
function.  Thus,  the  area  of  a  circle  and  its  radius  are  two 
related  quantities,  such  that  if  one  is  given,  the  other  is  deter- 
mined. We  can  say  that  the  area  is  a  function  of  the  radius, 
and  likewise  that  the  radius  is  a  function  of  the  area. 


10 


CARTESIAN  COORDINATES 


The  relation  between  the  independent  variable  and  the  function 
can  be  graphically  represented  by  the  use  of  rectangular  coordi- 
nates. For  if  we  represent  the  independent  variable  by  x  and  the 
corresponding  value  of  the  function  by  ?/,  x  and  y  will  determine 
a  point  in  the  plane,  and  a  number  of  such  points  will  outline  a 
curve  indicating  the  correspondence  of  values  of  variable  and 
function.    This  curve  is  called  the  graph  of  the  function. 

Ex.  1.  An  important  use  of  the  graph  of  a  function  is  in  statistical 
work.  The  following  table  gives  the  price  of  standard  steel  rails  per  ton 
in  ten  successive  years  : 

1895 $24.33  1900 $32.29 

1896 28.00  1901 27.33 

1897 18.75  1902 28.00 

1898 17.62  1903 28.00 

1899 28.12  1904 28.00 

If  we  plot  the  years  as  abscissas  (calling  1895  the  first  year,  1896  the 
second  year,  etc.)  and  plot  the  price  of  rails  as  ordinates,  making  one  unit 
of  ordinates  correspond  to 
ten  dollars,  we  shall  locate 
the  points  Pv  P2,  . . .,  P10  in 
fig.  18.  In  order  to  study  the 
valuation  in  price,  we  join 
these  points  in  succession  by 
straight  lines.  The  resulting 
broken  line  serves  merely  to 
guide  the  eye  from  point  to  Tig.  18 

point,    and    no    point    of    it 

except  the  vertices  has  any  other  meaning.  It  is  to  be  noted  that  there 
is  no  law  connecting  the  price  of  rails  with  the  year.  Also  the  nature  of 
the  function  is  such  that  it  is  defined  only 
for  isolated  values  of  x. 

Ex.  2.  As  a  second  example  we  take  the 
law  that  the  postage  on  each  ounce  or  frac- 
tion of  an  ounce  of  first-class  mail  matter  is 
two  cents.  The  postage  is  then  a  known 
function  of  the  weight.  Denoting  each  ounce 
of  weight  by  one  unit  of  x,  and  each  two 
cents  of  postage  by  one  unit  of  ?/,  we  have 
a  series  of  straight  lines  (fig.  19)  parallel 
to  the  axis  of  x,  representing  corresponding 

values  of  weight  and  postage.     Here  the  function  is  defined  by  United 
States  law  for  all  positive  values  of  x,  but  it  cannot  be  expressed  in 


$kQ 
$30 

Sjd 
$10 

T 

n 

i; 

n 

s       < 

/• 

1\ 

V.    1 

i:> 

O 

'9 

0        '9 

5    'I 

7    'i 

S   >9 

J    '0 

)    '0 

1    '0 

8    '0 

;   '0 

v     " 

] 

fir1 

r 

0 

lo 

z.    2 

->z.   3 

oz.    h 

oz.  * 

VARIABLE  AND  FUNCTION 


11 


elementary  mathematical  symbols.  A  peculiarity  of  the  graph  is  the  series 
of  breaks.  The  lines  are  not  connected,  but  all  points  of  each  line  repre- 
sent corresponding  values  of  x  and  y. 

Ex.  3.  As  a  third  example,  differing  in  type  from  each  of  the  preceding, 
let  us  take  the  following.  While  it  is  known  that  there  is  some  physical 
law  connecting  the  pressure  of  saturated  steam  with  its  temperature,  so 
that  to  every  temperature  there  is  some 
corresponding  pressure,  this  law  has  not 
yet  been  formulated  mathematically. 
Nevertheless,  knowing  some  correspond- 
ing values  of  temperature  and  pressure, 
we  can  construct  a  curve  that  is  of  con- 
siderable value.  In  the  table  below,  the 
temperatures  are  in  degrees  centigrade 
and  the  pressures  are  in  millimeters 
of  mercury. 


Temperature 

Pressure 

100 

7(50 

105 

906 

110 

1074.7 

115 

1208.7 

120 

1400.5 

125 

1743.3 

130 

2029.8 

135 

2853. 7 

140 

2717.9 

145 

3126.1 

150 

3581.9 

Y 

[_ 

i 

1 

1 

t 

_j 

1 

.    1 

f: 

_    7 

_  / 

r 

-1 

1 

r 

/ 

/ 

r 

1- 

L 

A 

t       - 

7        : 

o                i~- 

1.0"     ^ 

Let  100  represent  the  zero  point  of 
temperature,  and  let  each  unit  of  x  repre- 
sent 5  degrees  of  temperature;  also  let 
each  unit  of  y  represent  loo  millimeters 
of  pressure  of  mercury,  and  locate  the 
points  representing  the  corresponding 
values  of  temperature  and  pressure  given  Fig.  20 

in  the  above  table.    Through  the  points 

thus  located  draw  a  smooth  curve  (fig.  20);  that  is,  one  which  has  no 
sudden  changes  of  direction.  While  only  the  eleven  points  located  are 
exact,  all  other  points  are  approximately  accurate,  and  the  curve  may  be 
used  for  approximate  computation  as  follows :  Assume  any  temperature, 
and,  laying  it  off  as  an  abscissa,  measure  the  corresponding  ordinate  of 
the  curve.  While  not  exact,  it  will  nevertheless  give  an  approximate 
value  of  the  corresponding  pressure.    Similarly,  a  pressure  may  be  assumed 


12 


CARTESIAN  COORDINATES 


and  the  corresponding  temperature  determined.  It  may  be  added  that 
the  more  closely  together  the  tabulated  values  are  taken,  the  better 
the  approximation  from  the  curve;  but  the  curve  can  never  be  exact 
at  all  points. 

Ex.  4.  As  a  final  example,  we  will  take  the  law  of  Boyle  and  Mariotte 
for  perfect  gases;  namely,  at  a  constant  temperature  the  volume  of  a 
definite  quantity  of  gas  is  inversely  proportional  to  its  pressure.  It  follows 
that  if  we  represent  the  pressure  by 
x  and  the  corresponding  volume  by  //, 

then    )/  —  -,   where   k  is   a   constant 
x 

and  x  and  y  are  positive  variables. 
A  curve  (fig.  21)  in  the  first  quad- 
rant, the  coordinates  of  every  point 
of  which  satisfy  this  equation,  repre- 
sents the  comparative  changes  in 
pressure  and  volume,  showing  that 
as  the  pressure  increases  by  a  cer- 
tain amount  the  volume  is  decreased 
more  or  less,  according  to  the  amount 
of  .pressure  previously  exerted. 

This    example    differs    from    the 
1  (receding   in   that    the    law    of    the 

function  is  fully  known  and  can  be  expressed  in  a  mathematical 
formula.  Consequently  we  may  find  as  many  points  on  the  curve  as 
we  please,  and  may  therefore  construct  the  curve  to  any  required 
degree  of  accuracy. 

9.  Functional  notation.    When  y  is  a  function  of  x  it  is  cus- 
tomary to  express  this  by  the  notation 

y  =/0> 

Then  the  particular  value  of  the  function  obtained  by  giving  x 
a  definite  value  a  is  written  /(«).    For  example,  if 


Fig.  21 


f(p) 


3.c2  +  l, 


then 


/(2)=23+3.22+l=21, 
/(0)=08+3.02+l  =  l, 
/(_3)  =  (-3)»+3(-3)9  +  l  =  l, 
/0)=a8+3a2+l. 


PROBLEMS  13 

If  more  than  one  function  occurs  in  a  problem,  one  may  be 
expressed  as /(a;),  another  as  F(x),  another  as  </>(.r),  and  so  on. 
It  is  also  often  convenient  in  practice  to  represent  different 
functions  by  the  symbols  /x(«),  f.2(*),  /3(X)>  ete- 

Similarly,  a  function  of  two  or  more  variables  may  be  ex- 
pressed by  the  symbol  f(x,  y).  Then  /(a,  6)  represents  the 
result  of  placing  x—a  and  y—b  in  the  function.    Thus,  if 

f(x,  y)  =  a-2+3.///-4//J, 
/(a,  J)=aa+3a*-46a. 

PROBLEMS 

1.  Find  the  perimeter  of  the  triangle  the  vertices  of  which  are 
(3,  4),  (-  2,  4),  (2,  2). 

2.  Find  the  perimeter  of  the  quadrilateral  the  vertices  of  which 
are  (2,1),  (-2,  8),  (-6,  5),  (-2, -2). 

3.  Prove  that  the  triangle  the  vertices  of  which  arc  (—  3,  —  2), 
(1,  4),  (—  5,  0)  is  isosceles. 

4.  Prove  that  the  triangle  the  vertices  of  which  are  (— 1,  1); 
(1,  3),  (-  V3,  2  +  V3)  is  equilateral. 

5.  Prove  that  the  quadrilateral  of  which  the  vertices  are  (1,  3), 
(3,  6),  (0,  5),  (-  2,  2)  is  a  parallelogram. 

6.  Prove  that  the  triangle  (1,  2),  (3,  4),  (-  1,  4)  is  a  right 
triangle. 

7.  Prove  that  the   triangle   the   vertices   of  which  are   (2,  3), 
(_2,  5),  (-1,  -3)  is  a  right,  triangle. 

8.  Prove  that  (8,  0),  (0,  -  6),  (7,  -  7),  (1,  1)  are  points  of  a 
circle  the  center  of  which  is  (4,  —3).    What  is  its  radius? 

9.  Find  a  point  equidistant  from  (0,  0),  (1,  0),  and  (0,  2). 

10.  Find  a  point  equidistant  from  the  points  (—4,  3),  (4,  2),  and 

(1,-1)- 

11.  Find  a  point  equidistant  from  the  points  (1,  3),  (0,  6),  and 

(-  -*,  1). 

12.  Find  the  center  of  a  circle  passing  through  the  points  (0,  0), 
(4,  2),  and  (6,  4). 

13.  Find  a  point  on  the  axis  of  x  which  is  equidistant  from  (0,  5) 

and  (4,  2). 


14  CARTESIAN  COORDINATES 

14.  Find  the  points  which  are  5  units  distant  from  (1,  3)  and  4 
units  distant  from  the  axis  of  y. 

15.  A  point  is  equally  distant  from  the  points  (3,  5)  and  (—  2,  4), 
and  its  distance  from  OY  is  twice  its  distance  from  OX.  Find  its 
coordinates. 

16.  Find  the  slopes  of  the  straight  lines  determined  by  the  fol- 
lowing pairs  of  points  :  (1,  2),  (—  3,  1)  ;  (3,  —  1),  (—  5,  —  1)  ; 
(2,  3),  (2,  -  5). 

17.  Find  the  lengths  and  the  slopes  of  the  sides  of  a  triangle  the 
vertices  of  which  are  (3,  5),  (-  3,  2),  (5,  2). 

18.  A  straight  line  is  drawn  through  the  point  (5,  0),  having  the 
slope  of  the  straight  line  determined  by  the  points  (—  1,  2)  and 
(4,  —  2).    Where  will  the  first  straight  line  intersect  the  axis  OY? 

19.  One  straight  line  with  slope  —  \  passes  through  (2,  0),  and  a 
second  straight  line  with  slope  1  passes  through  (—  2,  0).  Where 
do  these  two  lines  intersect  ? 

20.  The  center  of  a  circle  with  radius  5  is  at  the  point  (1,  1). 
Find  the  ends  of  the  diameter  of  which  the  slope  is  J. 

21.  Two  straight  lines  are  drawn  from  (2,  3)  to  the  axis  OX.  If 
their  slopes  are  respectively  ^  and  —  2,  prove  that  they  are  the 
sides  of  a  right  triangle  the  hypotenuse  of  which  is  on  OX. 

22.  Find  the  coordinates  of  a  point  P  on  the  straight  line  deter- 
mined by  P  (-  2,  3)  and  P  (4,  6),  where  ^  =  -• 

1  1  2  O 

23.  A  point  of  the  straight  line  joining  the  points  (3,  —  1)  and 
(5,  —  5)  divides  it  into  segments  which  are  in  the  ratio  2  :  5.  What 
are  its  coordinates  ? 

24.  On  the  straight  line  determined  by  the  points  ^(4,  6)  and 
P2(—  2,  —  5)  find  the  point  three  fourths  of  the  distance  from 
Px  to  P2. 

25.  Find  the  points  of  trisection  of  the  line  joining  the  points 
Pj(-3,  -7)  and  P3(10,  2). 

26.  The  middle  point  of  a  certain  line  is  (—  1,  2),  and  one  end  is 
the  point  (2,  5).    Find  the  coordinates  of  the  other  end. 

27.  To  what  point  must  the  line  drawn  from  (1,  —1)  to  (4,  5)  be 
extended  in  the  same  direction,  that  its  length  may  be  trebled  ? 


P,  P       L 

Px(xv  y^  and  P.2(:'\,,  //.J,  such  that  ——  =  -  >  prove   that 


PKOBLEMS  15 

28.  One  end  of  a  line  is  at  (2,  —  3),  and  a  point  one  fifth  of  the 
distance  to  the  other  end  is  (1,  —  2).  Find  the  coordinates  of  the 
other  end. 

29.  Find  the  lengths  of  the  medians  of  the  triangle  (3,  4), 
(-1,1),  (0,-3). 

30.  The  vertices  of  a  triangle  are  ^4(0,  0),  B{—  2,  5),  and  C(4,  3). 
Show  that  the  slope  of  the  straight  line  joining  the  middle  points 
of  Ali  and  BC  is  the  same  as  the  slope  of  AC. 

31.  Find  the  slopes  of  the  straight  lines  drawn  from  the  origin 
to  the  points  of  trisection  of  the  straight  line  joining  (—  2,  4) 
and  (4,  7). 

32.  Given  the  three  points  A(—  2,  4),  P(4,  2),  and  C(7,  1)  upon 

a  straight  line.    Find  a  fourth  point  D,  such  that  '- —  =  —  '- — -• 
b  1  DC  BC 

33.  If  P(x,  y)  is  a  point  on  the  straight  line  determined  by 

P,  P      L 

y^),  such  that  ——  =  j  >  prov 

_  lyr.,  +  l.jrx  _  /,//,_,  +  /g//l 

34.  Given  four  points  J\,  /'.,,  Pp  I\.  Find  the  point  halfway 
between  P1  and  P2,  then  the  point  one  third  of  the  distance  from 
this  point  to  Pg,  and  finally  the  point  one  fourth  of  the  distance 
from  this  point  to  P4.  Show  that  the  order  in  which  the  points 
are  taken  does  not  affect  the  result. 

35.  Prove  analytically  that  the  lines  joining  the  middle  points 
of  the  opposite  sides  of  a  quadrilateral  bisect  each  other. 

36.  Prove  analytically  that  in  any  right  triangle  the  straight  line 
drawn  from  the  vertex  of  the  right  angle  to  the  middle  point  of  the 
hypotenuse  is  equal  to  half  the  hypotenuse. 

37.  Prove  analytically  that  the  straight  line  drawn  between  two 
sides  of  a  triangle  so  as  to  cut  off  the  same  proportional  parts,  meas- 
ured from  their  common  vertex,  is  the  same  proportional  part  of  the 
third  side. 

38.  OABC  is  a  trapezoid  of  which  the  parallel  sides  OA  and  CB 
are  perpendicular  to  OC.  D  is  the  middle  point  of  AB.  Prove 
analytically  that  OD  =  CD. 


16 


CARTESIAN    COORDINATES 


39.  Prove  analytically  that  the  diagonals  of  a  parallelogram  bisect 

each  other. 

40.  Prove  analytically  that  if  in  any  triangle  a  median  is  drawn 
from  the  vertex  to  the  base,  the  sum  of  the  squares  of  the  other  two 
sides  is  equal  to  twice  the  square  of  half  the  base  plus  twice  the 
square  of  the  median. 

41.  Prove  analytically  that  the  line  joining  the  middle  points  of 
the  nonparallel  sides  of  a  trapezoid  is  one  half  the  sum  of  the 
parallel  sides. 

42.  Prove  analytically  that  if  two  medians  of  a  triangle  are  equal, 
the  triangle  is  isosceles. 

43.  Show  that  the  sum  of  the  squares  on  the  four  sides  of  any 
plane  quadrilateral  is  equal  to  the  sum  of  the  squares  on  the  diagonals 
together  with  four  times  the  square  on  the  line  joining  the  middle 
points  of  the  diagonals. 

44.  The  following  table  gives  to  the  nearest  million  the  num- 
ber of  tons  of  pig  iron  produced  in  the  United  States  for  the  years 
indicated.    Eepresent  the  table  by  a  graph. 


1007 

25,000,000 

1011 

23,000,000 

1908 

10,000,000 

1912 

29,000,000 

1909 

25,000,000 

1913 

31,000,000 

1910 

27,000,000 

1914 

23,000,000 

45.  The  following  table  gives  to  the  nearest  thousand  the  number 
of  immigrants  into  the  United  States  during  the  year  1914.  Exhibit 
this  table  in  the  form  of  a  graph. 


January 

45,000 

July 

00,000 

February 

47,000 

August 

38,000 

March 

93,000 

September 

29,000 

April 

120,000 

October 

30,000 

May 

108,000 

November 

20,000 

June 

72,000 

December 

21,000 

PROBLEMS 


IT 


46.  The  average  yearly  precipitation  at  the  different  meteoro- 
logical stations  of  one  of  the  states  for  the  years  indicated  was  as 
follows.    Represent  the  table  by  a  graph. 


1905 

0.74  in. 

1910 

0.79  in. 

1000 

2.09  in. 

1911 

0.30  in. 

1907 

1.5(5  in. 

1912 

0.98  in. 

1908 

1.20  in. 

1913 

1.06  in. 

1909 

3.02  in. 

1914 

0.41  in. 

47.  The  daily  maximum  temperatures  at  a  certain  town  in  the 
United  States  for  the  first  ten  days  of  A.ugust,  L915,  were  respec- 
tively 100°,  99°,  93°,  89°,  92°,  94°,  96°,  95°,  '.Mi0,  97°.  Construct  a 
graph  showing  the  variation  in  temperature.  (Place  90°  at  the  zero 
point  of  the  temperature  scale.) 

48.  At  a  certain  place  the  readings  of  the  height  of  the  barometer 
taken  at  noon  and  midnight  for  a  week  were,  in  order,  as  follows: 
29.4,  29.6,  29.8,  30.1,  30.4*  30.8,  30.9,  30.4,  29.8,  29.3,  29.4,  29.7, 
29.6,  29.8.  Construct  a  graph  showing  the  changes  in  atmospheric 
pressure. 

49.  On  a  certain  Swiss  railroad,  the  stations  with  their  distance 
in  miles  from  the  first  station  and  their  elevation  in  feet  above 
sea  level  are  as  follows.  Represent  graphically  the  profile  of  the 
railroad. 


Station 

A 

B 

C 

I) 

E 

F 

0 

11 

/ 

J 

K 

Distance 

8  mi. 

9.5  mi. 

12mi. 

13mi. 

l.",ini. 

I8rai. 

20  mi. 

22.5  mi. 

25  mi. 

28mi. 

Elevation 

1437' 

1440' 

1580' 

1620' 

1555' 

L558 

1665' 

2305' 

216C 

3295 

I960' 

50.  In  a  test  on  the  tensile  strength  of  a  steel  rod,  originally  10  in. 
long,  subjected  to  a  varying  load,  the  following  readings  were  made, 
the  applied  load  being  expressed  in  pounds  per  square  inch  of  cross 
section  and  the  elongation  being  measured  in  inches.  Illustrate  the 
test  graphically. 


Load 

1000 

5000 

10,000 

20,000 

30,000 

(50,000 

65,000 

70,000 

Elongation 

0 

.0015 

.0031 

.0070 

.0108 

.0212 

.0230 

.0248 

18 


CARTESIAN  COORDINATES 


51.  A  varying  load,  expressed  in  pounds  per  square  inch  of  cross 
section,  was  applied  to  the  end  of  a  concrete  block  originally  5  in.  tall, 
and  the  corresponding  compression  was  measured  in  inches,  with  the 
results  expressed  in  the  following  table.  Illustrate  the  test  graphically. 


Load 

100 

200 

300 

400 

500 

600 

700 

800 

900 

1000 

Compression 

0 

.0004 

.0010 

.0021 

.0035 

.0054 

.0078 

.0107 

.0130 

.0178 

52.  A  body  is  thrown  vertically  upward  with  a  velocity  of  100  ft. 
per  second.  If  v  is  the  velocity  at  the  time  t,  v  =  100  —  gt.  Assuming 
g  =  32,  construct  the  graph  showing  the  relation  between  v  and  t. 

53.  The  space  s  through  which  a  body  falls  from  rest  in  a  time  t 
is  given  by  the  formula  s  =  \  gt2.  Assuming  g  =  32,  construct  the 
graph  showing  the  relation  between  s  and  t. 

54.  A  body  is  thrown  up  from  the  earth's  surface  with  an  initial 
velocity  of  100  ft.  per  second.  If  s  is  the  space  traversed  in  the 
time  t,  s  =  100 1  —  \  gt2.  Assuming  g  =  32,  construct  the  graph 
showing  the  relation  between  s  and  t. 

55.  Make  a  graph  showing  the  relation  between  the  side  and  the 
area  of  a  square. 

56.  Make  a  graph  showing  the  relation  between  the  radius  and 

the  area  of  a  circle. 

-p 

57.  Ohm's  law  for  an  electric  current  is  C  =  —  >  where  C  is  the 

current,  E  the  electromotive  force,  and  R  the  resistance.  Assuming 
E  to  be  constant,  plot  the  curve  showing  the  relation  between  the 
resistance  and  the  current. 

58.  Two  particles  of  mass  mx  and  m2  at  a  distance  d  from  each 
other  attract  each   other  with  a  force   F  given   by  the   equation 

F  =  — ^-^- •    Assuming   mx  =  5   and   m2  =  20,   construct   the    graph 

showing  the  relation  between  F  and  d. 

59.  If  f(x)  =  .t4  -  4  x2  +  6 x  - 1,  find /(3),  /(0),  /(-  2). 

60.  If  f{x)  =  xs  -  3  x2  +  1,  show  that/(2)  +  2/(0)  =/(l). 

61.  If  f(x)  =  x3  -  3  x2  +  5  x  -  6,  find  /(a),  /(-  a),  f(a  +  h). 

62-  If•^)  =  ^f^' 


find/(3),/(0),/(-l). 


PROBLEMS  19 

63.  If /(a-)  =  jA  ~ a(7"  +  '  ,  prove  that/(-  x)  =/(*). 

a.   -j-  A 

64.  If  /(a:)  =  xb  +  5  a;3  -  9  x,  prove  that  /(—«)  =  -  /(a;). 

65.  If  /(a-)  =  x2  +  2  aa:  -  a2,  prove  that  /(«)  +/(-  a)  =  0. 

66.  If  f(x)  =  (x  -  ^)  (x2  -  |) ,  prove  that /(a)  -/(£)■ 

67.  If /(a-)  =  |rff  >  prove  that  /(«.)  .  /(-  a)  =  1. 

68.  If/Car)  =  *4+g*'+5g  +  l>  prove  that /(a-)  =/(i)- 

69 .  If  fjx)  =  xz+  a*  and  f%(x)  =  2  ax,  prove  that  /, (a)-  «/,(«)  =  0. 

70.  If  fx(x)  =  Va;2  -  4  and  f2(x)  =  Vz2  +  4,  prove  that 

/.(•+D+/.(— D-*» 

71.  H/.(a:)=-\/-  +  \2  a=d  /,(*)=  \~  -  "\F>  P'«™  that 

[/1w]2-[/,(-'-)]2=[/1(")]2- 

72.  If  /(a-)  =  2y  +  1 ,  prove  that  /[/(.r)]  =  a-. 

73.  If /(a-,  y)=  x2  +  //  -  5,  find  /(0,  0),  /(l,  0),/(0,  1),/(1,  2) 

74.  If/(.r,  y)=^iy,  prove  that /(a,  &)  =  -/(&,  a). 

x      i) 

75.  If  /^aj,  y)  =  x  +  ij  and  f,(x,  y)  =  x  -  y,  prove  that 

76.  If f^  V)=^  +  l  and  f%{x,  y)  =  ^  +  £ ,  prove  that 

'    ./jr.  //)  =  [/;<,-,  y)]2- 2. 

77.  If  /^aj,  y)  =  a;  +  3  y  and  /2(a;,  y)  =  3  a:  +  9  //,  prove  that 

xA(x,y)+uf>(*,y)=[fi(*>y)T' 


CHAPTER  II 
GRAPHS  OF  ALGEBRAIC  FUNCTIONS 
10.  Equation  and  graph.   If  f(x)  is  any  function,  and  we  place 

*=/(*). 

we  may,  as  already  noted,  construct  a  curve  which  is  the  graph  of 
the  function.  The  relation  between  this  curve  and  the  equation 
y  =f(x)  is  such  that  all  points  the  coordinates  of  which  satisfy  the 
equation  lie  on  the  curve;  and  conversely,  if  a  point  lies  on  the 
curve,  its  coordinates  satisfy  the  equation. 

The  curve  is  said  to  be  represented  by  the  equation,  and 
the  equation  is  called  the  equation  of  the  curve.  The  curve 
is  also  called  the  locus  of  the  equation.  Its  use  is  twofold : 
on  the  one  hand,  we  may  study  a  function  by  means  of  the 
appearance  and  the  properties  of  the  curve ;  and  on  the  other 
hand,  we  may  study  the  geometric  properties  of  a  curve  by 
means  of  its  equation.  Both  methods  will  be  illustrated  in 
the  following  pages. 

Similarly,  any  equation  in  x  and  y  expressed  by 

represents  a  curve  which  is  the  locus  of  the  equation.  To 
construct  this  curve  we  have  to  find  enough  points  whose 
coordinates  satisfy  the  equation  to  outline  the  curve.  This 
may  be  done  by  assuming  at  pleasure  values  of  x,  substituting 
these  values  in  the  equation,  and  solving  for  the  corresponding 
values  of  y.  Before  this  computation  is  carried  out,  however, 
it  is  wise  to  endeavor  to  obtain  some  idea  of  the  shape  of  the 
curve.  The  computation  is  then  made  more  systematic,  or  in 
some  cases  the  curve  may  often  be  sketched  free-hand  with 
sufficient  accuracy. 

20 


INTERCEPTS 


21 


The  following  plan  of  work  is  accordingly  suggested: 

1.  Find  the  points  in  which  the  curve  intercepts  the  coordi- 
nate axes. 

2.  Find  if  the  curve  has  symmetry  with  respect  to  either  of 
the  coordinate  axes  or  to  any  other  line. 

3.  Fhid  if  any  values   of  one  variable  are   impossible,  since 
they  make  the  other  variable  imaginary. 

4.  Find  the  values   of  one  variable  which  make   the   other 
infinite. 

Each  of  the  above  suggestions  is  illustrated  in  one  of  the 
following  articles : 

11.  Intercepts.  The  curve  will  have  a  point  on  the  axis 
of  x  when  y  =  0  and  will  have  a  point  on  the  axis  of  y 
when  x  =  0.  Hence  we  may  find  the  intercepts  on  one  of  the 
coordinate  axes  by  placing  the  other 
coordinate  equal  to  zero  and  solving  the 
resulting  equation. 

Ex.  1.    y  =  .5  (./•  +  2)  (.-•  +  ..-,)  (.,•  -  2). 

If  y  =  0,  x=  —  2  or  —.5  or  2,  and  there  are 
three  points  of  the  curve  mi  the  axis  of  x. 

If  x  —  0,  y=—\,  and  there  is  one  intercept 
on  the  axis  of  //. 

If  x<—  2,  all  three  factors  are  negative; 
therefore  y  <  0,  and  the  corresponding  part  of 
the  curve  lies  below  the  axis  of  x.  If  —  2  <  x 
<—  .5,  the  first  factor  is  positive  and  the  other 
two  are  negative;  therefore  y  >  0,  and  the 
corresponding  part,  of  the  curve  lies  above  the 
axis  of  x.  If  —  .5  <  x  <  2,  the  first  two  factors 
are  positive  and  the  third  is  negative;  therefore 

y<0,  and  the  corresponding  part  of  the  curve  lies  below  the  axis  of  x. 
Finally,  if  x  >  2,  all  the  factors  are  positive;  therefore  //><»,  and  the 
corresponding  part  of  the  curve  lies  above  the  axis  of  x. 

Assuming  values  of  X  and  finding  the  corresponding  values  of  y,  we  plot 
the  curve  as  represented  in  fig.  22. 

Ex.  2.    y  =  .5  (x  +  2.5)  (x  -  l)2. 

If  y  =  0,  x  =  —  2.5  or  1,  and  there  are  two  points  of  the  curve  on  the 
axis  of  x. 

If  x  =  0,  y  =  1.25,  and  there  is  one  intercept  on  the  axis  of  ?/. 


Fig.  22 


22 


GRAPHS  OF  ALGEBRAIC  FUNCTIONS 


If    x  <  —  2.5,  the    first    factor   is    negative    and   the   second  factor  is 
positive  ;    therefore  y  <  0,  and  the  corresponding  part  of  the  curve  lies 
below  the  axis  of  x.    If  —  2.5  <  x  <  1,  both  factors  are  positive ;  therefore 
y  >  0,  and  the  corresponding  part  of  the  curve 
lies  above   the   axis  of  x.    Finally,  if  x  >  1,  we 
have  the  same  result  as  when  —  2.5  <  x  <  1,  and 
the  curve  does   not   cross  the   axis  of  x  at  the 
point  x  =  1  but  is  tangent  to  it. 

Assuming  values  of  x  and  finding  the  corre- 
sponding values  of  y,  we  plot  the  curve  as  repre- 
sented in  fig.  23. 

Since  it  will  be  shown  in  §  31  that  an 
equation  of  the  first  degree  in  x  and  ?/, 


Ax  +  By+  C=Q, 


Fig.  23 


always  represents  a  straight  line,  and  since  a  straight  line  is 
determined  by  two  points,  it  is  generally  sufficient  in  plotting 
an  equation  of  the  first  degree  to  find  the  intercepts  on  the 
two  axes  and  draw  a  straight  line  through  the  two  points  thus 
determined.  The  only  exception  is  when  the  straight  line  passes 
tlrrough  the  origin,  in  which  case  some  point  of  the  straight 
line  other  than  the  origin  must  be  found  by  trial. 

Ex.  3.  Plot  the  line  3a;  —  5y  +  12  =  0.  Placing  y  =  0,  we  find  x  =  —  4. 
Placing  x  =  0,  we  find  y  =  2f.  We  lay  off  OL  =  -  4,  OK  =  2|,  and  draw  a 
straight  line  through  L  and  K  (fig.  24). 


Fig.  24 


Fig.  25 


Ex.  4.    Plot  the  line  3  x  —  5  y  =  0.    Here,  if  x  =  0,  y  -  0.    If  we  place 
=  1,  we  find  y  =  f .    The  line  is  drawn  through  (0,  0)  and  (1,  f)  (fig.  25). 


SYMMETRY 


23 


12.  Symmetry  and  impossible  values.  A  curve  is  symmetri- 
cal with  respect  to  the  axis  of  x  when  to  each  value  of  x 
in  its  equation  correspond  two  values  of  y,  equal  in  magni- 
tude and  opposite  in  sign.  This  occurs  in  the  simplest  manner 
when  y  is  equal  to  plus  or  minus  the  square  root  of  a  func- 
tion of  x.  Any  value  of  x  which  makes  the  function  under 
the  radical  sign  positive  gives  two  points  of  the  curve  equi- 
distant from  the  .r-axis.  Values  of  x  which  make  the  function 
under  the  radical  sign  negative  make  y  imaginary  and  give 
no  points  of  the  curve.  These  values  of  x  we  call  impossible 
values. 

Similar  remarks  hold  for  symmetry  with  respect  to  the  axis 
of  y.  How  symmetry  with  respect  to  other  lines  may  sometimes 
be  determined  is  shown  by  Ex.  5. 


Ex.1.    y=±  V(x  +  2)  (x  -  1)  (x  -  5). 

If  x=—  2,  1,  or  5,  y  =  0,  and  the   graph    intersects   the   axis   of   X  at 
three   points. 

The  lines  x=—  2,  x  —  1,  and  x  =  5  divide  the  plane  (fig.  26)  into  four 
sections. 

If  x  <  —  2,  all  three  factors  of  the 
product  are  negative;  hence  the  radi- 
cal is  imaginary  and  there  can  be  DO 
part  of  the  graph  in  the  correspond- 
ing section  of  the  plane.    If  —  2<x<l, 

the  first  factor  is  positive  and  the 
other  two  are  negative;  hence  the 
radical  is  real  and  there  is  a  pari  of 
the  graph  in  the  corresponding  section 

of  the  plane.  If  1  <  x  <  .">,  the  first 
two  factors  are  positive  and  the  third 
is  negative;  hence  the  radical  is 
imaginary  and  there  can  be  no  part 
of  the  graph  in  the  corresponding' 
section  of  the  plane.  Finally,  if 
x  >  5,  all  three  factors  are  positive ; 
hence  the  radical  is  real  and  there 
is  a  part  of  the  graph  in  the  corre- 
sponding section  of  the  plane. 

Therefore  the  graph  consists  of  two  separate  parts  and  is  seen  (fig.  2G) 
to  consist  of  a  closed  loop  and  a  branch  of  infinite  length. 


24 


GRAPHS  OF  ALGEBRAIC  FUNCTIONS 


Ex.  2.    y  =  ±  V(jc  +  2) (a;  -  l)8. 

This  will   be  written  as 

i/  =  ±(x-l)Vx  +  2. 

The  line  x  =  —  2  divides  the  plane  (fig.  27) 
into  two  sections. 

Proceeding  as  in  the  previous  example,  we 
find  the  radical  to  be  real  if  x  >  —  2  and 
imaginary  if  x  <  —  2.  Therefore  there  is  a 
part  of  the  graph  to  the  right  of  the  line 
x  =  —  2,  but  there  can  be  no  part  of  the 
graph  to  the  left  of  that  line  unless  x  can 
have  a  value  that  makes  the  coefficient  of  the 
radical  zero;  and  this  coefficient  is  zero  only 
when  x  equals  unity.  Hence  all  of  the  graph  lies  to  the  right  of  the 
line  x  =  —  2,  as  shown  in  fig.  27. 

To  every  value  of  x  correspond  two  values  of  y  which  are  in  general 
distinct  but  become  equal  when  x  =  1.  Hence  the  curve  crosses  itself 
when  x  =  1. 

Comparing  this  example  with  Ex.  1,  we  see  that  by  changing  the 
factor  x  —  5  to  x  —  1  we  have  joined  the  infinite  branch  and  the  loop, 
making  a  continuous  curve  crossing  itself  at  the  point  (1,  0). 


Fig.  27 


Ex.  3,    y  =  ±V(x  +  2)2(x-l) 

=  ±(x  +  2)Vx-l. 

The  line  x  = 1  divides  the  plane  (fig.  28)  into 
two  sections. 

If  x  >  1,  the  radical  is  real  and  there  is  a 
part  of  the  graph  in  the  corresponding  sec- 
tion of  the  plane.  If  x  <  1,  the  radical  is 
imaginary  and  there  will  be  no  points  of  the 
graph  excel >t  Ior  sucn  values  of  x  as  make 
the  coefficient  of  the  radical  zero.  There  is 
but  one  such  value,  —  2,  and  therefore  there 
is  but  one  point  of  the  graph,  (—  2,  0),  to 
the  left  of  the  line  x  =  1.  The  graph  con- 
sists, then,  of  the  isolated  point  A  and  the 
infinite   branch   (fig.  28). 

Comparing  this  example  also  with  Ex.  1, 
we  see  that  by  changing  the  factor  x  —  5  to 
x  +  2  we  have  reduced  the  loop  to  a  single 
point,   leaving  the   infinite  branch   as   such. 


INFINITE  VALUES 


25 


Ex.4.    y  =  ±V-(x  +  4:)(x  +  2)2(x-fy 
=  ±  (x  +  2)  V-(a;  +  4)(x-4). 

The  lines  x  =  —  4  and  x  =  4  divide  the  plane 
(fig.   29)   into  three   sections. 

If  —  4  <  x  <  4,  the  radical  is  real  and  there  is  a 
part  of  the  graph  in  the  corresponding  portion  of 
the  plane.  If  x<—  4  or  x  >  4,  the  radical  is  imagi- 
nary; and  since  in  the  corresponding  sections  there 
is  no  value  of  x  which  "makes  x  +  2  zero,  there  can 
be  no  part  of  the  graph  in  those  sections.  It  is 
represented   in   fig.   29. 

Ex.  5.    2  x"  +  f  +  3  x  -  4  y  -  5  =  0. 

Solving  fur  y,  we  have 

2  ±y/~. 


y 


3  x  +  9, 


Fig.  29 


or,    after    the    expression     under    the 

radical  sign   has  been   factored, 


y  =  2±V-2(*-|)(*  +  8> 

The  lines  x  =  —  3  and  x  =  •?,  divide 
the  plane  (fig.  30)  into  three  sections, 
and  proceeding  as  before,  we  find  that 
the  curve  is  entirely  in  the  middle 
section  (that  is,  when  —  3  <x  <  §) 
and  that  the  line  y  —  2  is  an  axis  of 
symmetry. 

If  now  we  should  solve  for  X  in 
terms  of  ?/,  we  should  find  another 
axis  of  symmetry,  a?  =  —  |,  and  that 
the  curve  is  bounded  by  the  lines 
y  =  —  1.2    and   y  =  5.2. 


ii !  /           In 

i 

\ 

'    \          !   ° 

1  z/=->-~      1 

i 
i 

Fig.  30 


13.  Infinite  values.  If  the  expression  defining  a  function  con- 
tains fractions,  the  function  is  not  defined  for  a  value  of  x  which 
makes  the  denominator  of  any  fraction  zero.  But  if  x  =  a  is  a 
value  which  makes  the  denominator  zero,  but  not  the  numerator, 
and  if  x  is  allowed  to  approach  a  as  a  limit,  the  value  of  the  func- 
tion increases  indefinitely  and  is  said  to  become  infinite.  The 
graph  of  a  function  then  runs  up  or  down  indefinitely,  approach- 
ing the  line  x  —  a  indefinitely  near  but  never  reaching  it. 


20 


GRAPHS  OF  ALGEBRAIC  FUNCTIONS 


This  is  expressed  concisely  by  the  formula 


Other  important  formulas  involving  infinity 


GO  X   G  =  00, 


which  may  be  explained  in  a  similar  manner.    For  example,  to 

c  c 

obtain  the  meaning  of  — ,  we  may  write  -  and  then  allow  x  to 

oo  x 

increase  indefinitely.    It  is  obvious  that  the  quotient  decreases  in 
numerical  value  and  may  be  made  as  small  as  we  please  by  tak- 


ing x  large  enough.    This  is  the  meaning  of  the  formula  —  =  0. 

00 

It  is  evident  that  y  is  real  for  all  values  of  x ;  also,  if  x  <  2,  y  is  nega- 

y  is 


tive,  and  if  x  >  2,  y  is  positive.    Moreover,  as  x  increases  toward 
negative  and  becomes  indefinitely  great ;  while  as  x  decreases  toward  2,  y  is 
positive  and  becomes  indefinitely  great.    We 
can  accordingly  assign  all  values  to  x  except  2. 
The  curve  is  represented  in  fig.  31. 

It  is  seen  that  the  nearer  to  2  the  value 
assigned  to  x,  the  nearer  the  corresponding 
point  of  the  curve  to  the  line  x  =  2.  In  fact, 
we  can  make  this  distance  as  small  as  we 
please  by  choosing  an  appropriate  value  for  x. 
At  the  same  time  the  point  recedes  indefi- 
nitely from  OX  along  the  curve. 

Now,  token  a  straight  line  has  such  a  position 
with  respect  to  a  curve  that  as  the  two  are 
indefinitely  prolonged  the  distance  between  them 
approaches  zero  as  a  limit,  the  straight  line  is 
called  an  asymptote  of  the  curve. 

It  follows  from  the  above  definition  that  the  line  x  =  2  and  also  the 
line  y  =  0  are  asymptotes  of  this  curve.  In  this  example  it  is  to  be  noted 
that  the  asymptote  x  =  2  is  determined  by  the  value  of  x  which  makes 
the  function  infinite. 

It  is  clear  that  all  equations  of  the  type 

1 


represent  curves  of  the  same  general  shape  as  that  plotted  in  fig.  31. 


INFINITE  VALUES 


27 


Ex.  2. 


+  2 


If  x 


2   or  if   x 


is  infinite;  hence  these  two 
values  may  not  be  assigned 
to  x,  all  other  values,  however, 
being  possible.  The  curve  is 
represented  in  fig.  32. 

By  a  discussion  similar 
to  that  of  Ex.  1  it  may  be 
proved  that  the  lines  x  =  —  2 
and  x  —  2,  which  correspond 
to  the  values  of  x  which  make 
the  function  infinite,  and  also 
the  line  y  =  0,  are  asymptotes 
of  the  curve. 

This  curve  is  a  special  case 
of  that  represented  by 


1 


"  c> 

II 

II 

H 

« 

V. 

"1 

0 

1 

Fig.  32 


and  it  is  not  difficult  to  see  how  the  curve  r< -presented  by 

X  —  (l        J—  h        X  —  C 


will  look  for  any  number  of  terms. 


Ex.  3. 


(*-2)s 


All  values  of  x  may  be  assumed  except  2. 
The  curve  is  represented  in  fig.  :!:'>.  It  is 
evident  that  the*  lines  x  —  2  and  ij  =  0  are 
asymptotes. 

This  curve  is  a  special  case  of  that  repre- 
sented by 

1 
y     (x-af 

which  is  itself  a  special  case  of 


(x_a)2      (x_6)a 


Fig.  33 


28 


GKAPHS  OF  ALGEBltAIC  FUNCTIONS 


Ex.  4.    f 


1 
a- 3' 


We  solve  for  y,  forming'  the  equation  y  =  ±  ■ 


The  line  x  = 


(fig.  34)  divides  the  plane  into  two  sections,  and  it  is  evident  that  there 
can  be  no  part  of  the  curve  in 
that    section    for    which    x  <  3. 
Moreover,  this  line   x  =  3   is  an 
asymptote,  as  in    the    preceding 
examples.    The  curve,  which  is  a 
special  case  of  that  represented  by 
1 
x  —  a 
is  represented  in  fig.  34.    It  is  to 
be  noted  that  the  axis  of  x  also 
is  an  asymptote. 


f 


Ex.  5.   y  = 


a;2  +  1 


To  plot  this  curve  we  write  the  equation  in  the  equivalent  fori: 
,  =  ,  +  1. 


(1) 


It  is  evident  that  all  values  except  0  may  be  assigned  to  x,  that  value 
being  excluded  as  it  makes  y  infinite.    Let  us  also  draw  the  line 

y  =  x,  (2) 

a  straight  line  passing  through  the  origin 
and  bisecting  the  first  and  the  third 
quadrants. 

Comparing  equations  (1)  and  (2),  we 
see  that  if  any  value  xx  is  assigned  to  x, 
the  corresponding  ordinates  of  (1)  and  (2) 


a?i  + 


and 


and  that 


Moreover,  the  numerical 


are  respectively 

they  differ  by  —  • 

value  of  this  difference  decreases  as  greater 

numerical  values  are  assigned  to  xv  and 

it  can  be  made   less  than   any  assigned 

quantity    however    small    by    taking    x1  Fig.  35 

sufficiently    great.     It    follows    that    the 

line  y  =  x  is  an  asymptote  of  the  curve.    It  is  also  evident  that  the  line 

x  =  0,  determined  by  the  value  of  x  which  makes  the  function  infinite, 

is  an  asymptote.    The  curve  is  represented  in  fig.  35. 


INTERSECTION  29 

14.  Intersection  of  graphs.    Let 

/„<>,£>  =0  (1) 

and  /„(«,y)=0  (2) 

be  the  equations  of  two  curves.  It  is  evident  that  any  point 
common  to  the  two  curves  will  have  coordinates  satisfying  both 
(1)  and  (2),  and  that,  conversely,  any  values  of  x  and  y  which 
satisfy  both  (1)  and  (2)  are  coordinates  of  a  point  common  to 
the  two  curves.  Hence,  to  find  the  points  of  intersection  of  two 
curves,  solve  their  equations  simultaneous///. 

The  simplest  case  which  can  occur  is  that  where  each 
equation  is  of  the  first  degree  and  hence  (§  31)  represents 
a  straight  line.  In  general  there  is  a  single  solution,  which 
locates  the  single  point  of  intersection  of  the  two  straight 
lines.  If  no  solution  can  be  found,  it  is  evident  that  the  lines 
are  parallel. 

Other  important  cases  are  the  two  following: 

Case  I.  f(x,  y)  =  0  and  fn(x,  y)  =  0.     Let 

/1(^^)=0'  (1) 

/„O>30=0,  (2) 

be  a  linear  equation  and  an  equation  of  the  nth  degree,  where 
n  >  1.    The  degree  of  a  curve  is  defined  as  equal  to  the  degree 

of  its  equation.  Accordingly  this  problem  is  to  find  the  points 
of  intersection  of  a  straight  line  and  a  curve  of  the  rath  degree, 
and  the  method  of  solution  is  as  follows: 

Solve  (1)  for  either  x  or  y  and  substitute  the  result  in  (2). 
If,  for  example,  we  solve  (1)  for  ?/,  the  result  of  substituting 
this  value  in  (2)  will  in  general  be  an  equation  of  the  rath 
degree  in  x,  the  real  roots  of  which  are  the  abscissas  of  the 
required  points  of  intersection.  The  ordinates  of  the  points  of 
intersection  are  now  found  by  substituting  in  succession  in  (1) 
the  values  of  x  which  have  been  found. 

If  two  roots  x1  and  x2  of  the  equation  in  x  are  equal,  the  cor- 
responding  ordinates   are  equal   and  the  two  points   coincide. 


30 


GRAPHS  OF  ALGEBRAIC  FUNCTIONS 


We  may  regard  this  case  as  a  limiting  case  when  the  position 
of  the  curves  is  changed  so  as  to  make  xx  and  x2  approach  each 
other ;  that  is,  so  as  to  make  the  points  of  intersection  of  the 
straight  line  and  the  curve  approach  each  other  along  the 
curve.  Accordingly  the  straight  line  represented  by  equation 
(1)  is,  by  definition,  tangent  to  the  curve  represented  by  equa- 
tion (2).  In  general  the  tangent  line  simply  touches  the 
curve,    without    cutting    it,    as    in   the    case    of   the   circle. 


Ex.  1. 

Find 

the 

jioints   of 

intersection  of 

3a; 

-2y- 

-4  = 

0 

(1) 

and 

x"  — 

iy  = 

0. 

(2) 

Solving  (1)  for  y  and  sub- 
stituting the  result  in  (2),  we 
have  x2  —  G  x  +  8  =  0,  the  roots 
of  which  are  2  and  4.  Substi- 
tuting these  values  of  x  in  (1), 
we  find  the  corresponding 
values  of  y  to  be  1  and  4. 
Therefore  the  points  of  inter- 
section are  (2,  1)  and  (4,  4) 
(fig.  36). 


Fig. 


Ex.  2.  Find  the  points  of 
intersection  of 

0  x  -  4  y  -  9  =  0  (1) 

and  x2  -  4  y  =  0.         (2) 

Solving  (1)  for  y  and  sub- 
stituting the  result  in  (2),  we 
have  x2  -  G  x  +  9  =  0.  The 
roots  of  this  ecpiation  are 
equal,  each  being  3.  Hence 
the  straight  line  is  tangent 
to  the  curve.  Substituting  3 
for  x  in  (1),  we  find  y  =  Z ; 
hence  the  point  of  tangency 
is  (3,  |)  (fig.  37). 


Fig.  37 


INTERSECTION 


31 


Ex.  3.    Find  the  points  of 
intersection  of 


3x-2v/-5  =  0 
and  x2  —  4  y  =  0. 


(1) 
(2) 


Proceeding  as  in  the  two 
previous  examples,  we  obtain 
x2  -  6  x  + 10  =  0,  the  roots  of 
which  are  3  ±V—  1.  Hence 
the  straight  line  does  not 
intersect  the  curve  (fig.  38). 
The  corresponding  values  of 
y  are  2  ±  %  V^l. 

Ex.  4.  Find  the  points  of 
intersection  of 


y  =  2x  (1) 

and  y2  =  x(x-Z)2.    (2) 

Substituting   the    value    of    y   from    (1)    in    (2), 
we    have 

ar[(x-3)a-4z]  =  0, 

or  x[x2-10x  +  !)]  =  0. 

Its  roots  are  0,  1,  and  9.  The 
corresponding  values  <>f  y  are 
found  from  (1)  to  be  0,  2, 
and  18.  Therefore  the  points 
of  intersection  are  (0,  0), 
(1,  2),  and  (9,  18)   (fig.  39). 

Ex.  5.    Find  the  points  of 
intersection  of 


Fig.  40 


y  =  3  x  +  2         (1) 

and  y  =  xs.  (2) 

Substituting  in  (2),  we  have 
Xs  -  3  x  -  2  =  0, 
or      (.r-2)(x  +  l)2  =  0. 

Its  roots  are  2,  —1,  —  1.  The  corresponding  values 
of  y,  found  from  (1),  are  8,  —  1,  —  1.  Therefore  the 
points  of  intersection  are  (2,  8)  and  (—  1,  —  1),  the 
latter  being  a  point  of  tangency  (fig.  10). 


Fig.  39 


32 


GRAPHS  OF  ALGEBRAIC"  FUNCTIONS 


12). 


(1) 
(2) 


Ex.  6.    Find  the  points  of  intersection  of 
2x  +  y-4  =  0 
and  if-  —  x  (a? 

After  substitution  we  have  x3  —  4  x2  +  4  x  —  16  =  0,  or  (x  —  4)  (x2  +  4)  =  0, 
the  roots  of  which  are  4  and  ±  2  V—  1.  The  corresponding  values  of  y, 
found  from  (1),  are  -  4  and  4  =F  4  V^.  The 
only  real  solution  of  equations  (1)  and  (2) 
being  x  —  4  and  y  =  —  4,  the  straight  line 
and  the  curve  intersect  in  the  single  point 
(4,  -  4)  (fig.  41). 

Case  II.  fm(x,  y)  =  0  and  fn(x,  z/)  =  0. 

Let  /mO,  #)  =  0  (1) 

be  an  equation  of  the  with  degree,  and 

/„£*,  y)  =  0  (2) 

be  an  equation  of  the  nth.  degree,  where 
m  and  n  are  both  greater  than  unity. 

The  method  is  the  same  as  in  the  preceding  case ;  that  is, 
the  elimination  of  either  x  or  y,  the  solution  of  the  resulting 
equation,  and  the  determination  of  the  corresponding  values 
of  the  unknown  quantity  eliminated.  The  equation  resulting 
from  the  elimination  is  in  general  of  degree  7nn,  and  the 
number  of  simultaneous  solutions  of  the  original  equations 
is  mn.  If  all  these  solutions  are  real  and  distinct,  the  corre- 
sponding curves  intersect  at  mn  points.  If,  however,  any  of 
these  solutions  are  imaginary,  or  are  alike  if  real,  the  correspond- 
ing curves  will  intersect  at  a  number  of  points  less  than  mn. 
Hence  two  curves  of  degrees  m  and  n  respectively  can  intersect  at 
mn  points  and  no  more. 


Tig.  41 


and 


Ex.  7.    Find  the  points  of  intersection  of 

=  0 
=  0. 


y 

x2  +  y" 


_  o 


(1) 
(2) 


Subtracting  (1)  from  (2),  we  eliminate  y,  thereby  obtaining  the  equa- 
tion x2  +  2  x  —  8  =  0,  the  roots  of  which  are  —  4  and  2.     Substituting  2 


INTERSECTION 


33 


anrl  _  4  in  either  (1)  or  (2),  we  find  the  corresponding  values  of  y  to  be 
±  2  and  ±  2  V—  2.    The  real  solutions  of  the  equations  are  accordingly 
x  =  2,  y  =  ±  2,  and  the  corresponding 
curves   intersect  at  the  points  (2,  2) 
and  (2,  -  2)  (fig.  42). 

From  the  figure  it  is  also  evident 
that  the  value  —  4  for  x  must  make  y 
imaginary,  as  both  curves  lie  entirely 
to  the  right  of  the  line  x  =  —  4. 

Ex.  8.  Find  the  points  of  intersec- 
tion of 


and 


3  y 

3  a: 


(1) 
y*  -  6  x  =  u.  (2) 

Substituting  in  (2)  the  value  of  y 
from  (1),  we  haw  x4 -  27j;  =  0.  This 
equation  may  be  written 

s(z-3)(x*  +  8x  +  9)'  =  0, 

the  roots  of  which  are  0,  3,  and — 


V-3 


Fig.  42 


Substituting  these  values 


of  x  in  (1),  we    find   the   corresponding  values  of   y  to   be  0,  :'>.  and 


3T3V-  :$ 


Therefore  the  real  solutions  of  the 


ationa  are  x  =  0, 


y  =  0  and  x=  3,  y=  '■'>.  If  we  had 
substituted  the  values  of  x  in  (2), 

we  should  have  at  first  seemed  to 
find  an  additional  real  solution, 
y  =—  3  when  a:  =  :>.  But  —  3  for 
y  makes  x  imaginary  in  (1 ),  as  no 
part  of  (1)  is  below  the  axis  of  x. 
Geometrically,    the    line    x  =  -\ 

intersects  the  curves  (1)  and  (2) 
in  a  common  point  and  also 
intersects  (2)  in  another  point, 
Therefore  the  only  real  solutions 
of  these  equations  are  the  ones 
noted  above,  and  the  correspond- 
ing curves  intersect  at  the  two 
points  (0,  0)  and  (3,  3)  (fig.  43). 

We  see,  moreover,  that  any  results  found  must  he  tested  by  substitution  in  both 
of  the  original  equations. 

The   remaining   two    solutions    of    these    equations,    found    by   letting 

-  3  ±  3  V^3 

x  = >  are  imaginary. 


34 


GRAPHS  OF  ALGEBRAIC  FUNCTIONS 


Ex.  9.    Find  the  points  of  intersection  of 

2  x2  +  3  y*  =  35  (1) 

and  xy  =  6.  (2) 

Since  these  equations  are  homogeneous  quadratic  equations,  we  place 

y  =  mx  (3) 

and  substitute  for  y  in  both  (1)  and  (2).   The  results  are  2  a;2  +  3  ra2x2  =  35 


and  mx2  =  6,  whence 

x2-        35 

(4) 

2  +  3  m2 

and                  x2  =  —  • 
m 

(5) 

Therefore 

35             6 

«\\ 

2  +  3  m2       m  v  ' 

from  which  we  find  m  =  %  or  f . 
If  m=  i)  then,  from  (5),  x=±2; 
and  from  (3)  the  corresponding 
values  of  y  are  ±  3. 

If  m  =  §,  in  like  manner  we 
find  x  =  ±  \  ^Q  and  y  =  ±  1 V6. 

Therefore  the  curves  intersect  at  the  four  symmetrically  situated 
points    (2,  3),   (-  2,    -  3),    (3  Vo",    2  Vo),    (-  1  Vti,    -  §  Vo)    (fig.  44). 

Ex.  10.    Find    the    points 
of   intersection   of 

2y2  =  x-2      (1) 
and       x2  -  4  y2  =  4.  (2) 

Eliminating  ?/,  we  have 

x2  -  2  x  =  0, 

the    roots    of    which    are    0 

and     2.      When     x  =  0     we 

find   from   either  (1)  or  (2) 

y  —  ±  V—  1,  and  when  x  =  2  either  (1)  or  (2)  reduces  to  y2  =  0,  whence 

y  =  0.   Therefore  these  two  curves  are  tangent  at  the  point  (2,  0)  (fig.  45). 

15.  Real  roots  of  an  equation.  It  is  evident  that  the  real 
roots  of  the  equation  f(x)  =  0  determine  points  on  the  axis  of 
x  at  which  the  curve  y  =f(x)  crosses  or  touches  that  axis. 
Moreover,  if  xx  and  x„  (xx  <  x2~)  are  two  values  of  x  such  that 


X 


Fig.   it 


REAL  ROOTS  35 

f(x^)  and  /(#2)  are  of  opposite  algebraic  sign,  the  graph  is  on 
one  side  of  the  axis  when  x  =  x^  and  on  the  other  side  when 
x  =  x2.  Therefore  it  must  have  crossed  the  axis  an  odd  number 
of  times  between  the  points  x  =  xx  and  x  =  x2.  Of  course  it  may 
have  touched  the  axis  at  any  number  of  intermediate  points. 
Now  if  f(x)  has  a  factor  of  the  form  (x—  a~)k,  the  curve  y=f(x) 
crosses  the  axis  of  x  at  the  point  x  =  a  when  k  is  odd,  and 
touches  the  axis  of  x  when  k  is  even.  In  each  case  the  equa- 
tion f(x)  =  0  is  said  to  have  k  equal  roots,  x  =  a.  Since  then  a 
point  of  crossing  corresponds  to  an  odd  number  of  equal  roots 
of  an  equation  and  a  point  of  touching  corresponds  to  an  even 
number  of  equal  roots,  it  follows  that  the  equation  f(x)  =  0 
has  an  odd  number  of  real  roots  between  xx  and  #2,  if  /(',) 
and  f(x^)  have  opposite  signs. 

The  above  gives  a  ready  means  of  locating  the  real  roots  of 
an  equation  in  the  form/(.r)=  0,  for  we  have  only  to  find  two 
values  of  #,  as  xx  and  x2,  for  which  f(x)  has  different  signs.  We 
then  know  that  the  equation  has  an  odd  number  of  real  roots 
between  these  values,  and  the  nearer  together  x  and  x  ,  the  more 
nearly  do  we  know  the  values  of  the  intermediate  roots.  In  locat- 
ing the  roots  in  this  manner  it  is  not  necessary  to  construct  the 
corresponding  graph,  though  it  may  be  helpful. 

Ex.  Find  a  real  root  of  the  equation  i8  +  2/-  17  =  0,  accurate  to  two 
decimal  places. 

Denoting  xs  +  2  x  —  17  by /"(•>')  al"^  assigning  successive  integral  values 
to  x,  we  nnd/(2)=  —  5  and/(3)  =  lG.  Hence  then'  is  a  real  root  of»the 
equation  between  2  and  3. 

We  now  assign  values  to  x  between  2  and  3,  at  intervals  of  one  tenth, 
as  2.1,  2.2,  2.3,  etc.,  and  we  begin  with  the  values  nearer  2,  since;  /(2)  is 
nearer  zero  than  is  /(3).  Proceeding  in  this  way  we  find/(2.3)  =—  .233 
and/(2.4)  =  1.624  ;  hence  the  root  is  between  2.3  and  2.1. 

Now,  assigning  values  to  x  between  'J.:;  and  2.4  at  intervals  of  one  hun- 
dredth, we  find  /(2.31)=—  .054  and /(2.32)  =  .127 ;  hence  the  root  is 
between  2.31  and  2.32. 

To  determine  the  last  decimal  place  accurately,  we  let  x  =  2.315  and 
find  /(2.315)  =  .037.  Hence  the  root  is  between  2.31  and  2.315  and  is 
2.31,  accurate  to  two  decimal  places. 

If  /(2.315)  had  been  negative,  we  should  have  known  the  root  to  be 
between  2.315  and  2.32  and  to  be  2.32,  accurate  to  two  decimal  places. 


36  GRAPHS  OF  ALGEBRAIC  FUNCTIONS 

PROBLEMS 

Plot  the  graphs  of  the  following  equations: 

1.  3* +  4// -7=0.  18.   //=(.r  +  l)(.r-4)(.x-3)2. 

2.  2x-5y+G  =  0.  19.  i/=(x-l)(x  +  3)(x*  +  2). 

3.  a-4-7y  =  0.  20.   y=(x-l)2(2x24-6a:  +  5). 

4.  4:r  — 3  =  0.  21.   y2  =  (x  -  2)  (a;2  -  9). 

5.  3  y  +  8  =  0.  22.  y2  =  {x  +  3)(6  a  -  a:2  -  8). 

6.  y  =  4a2+4a-3.  23.   9  y2  =  (x  +  2)  (2  x  -  If. 

7.  y  =  4  sr  -  2  x  +  3.  24.   4  y2  =  a;3  +  4  a:2. 

8.  y  =  6  -  x  -  x2.  25.9  //  =  (a2  - 1)  (4  x2  -  25). 

9.  7/  =  -  3  x2  +  4  x.  26.  y2  =  (1  -  a2)  (4  a;2  -  25). 

10.  y  =  (a4-2)(.r-l)(.r-3).  27.   4  y2  =  9  xi  -  a6. 

11.  y=  (ft-2- 1)  (2.x  +  9).  28.  y2  =  (2a  +  3)(4a2-9). 

12.  y  =  a3  +  4  a-2.  29.   y2  =  (x  -  2)2(3  -  2  a;). 

13.  y  =  (x-3)(2x  +  lf.  30.   y2=(2  +  a-a2)fa  +  2)2. 

14.  y  =  a;8-8a;a+15aj.  31.  f  =  x2 (x  -  5)a(2 x  -  3). 

15.  y  =  2  ft3  +3  a2  -14  «.  32.   4y2=(a--l)2(4a2-4a-3). 

16.  y  =  a8  -  a2  -  4  a;  +  4.  33.  f  =  a  (as  +  2)2(x  +  3)2. 

17.  y  =  xs-  a2x.  34.  y2  =  (2  a;  -  3)  (a2  + 1). 

35.  y2  =  (x  - 1)  (2  x  -  l)2(.r2  -f  3 x  +  3). 

36.  a-2  4-  y2  —  4  a;  4-  6  y  +  9  =  0. 

37.  K2_4y_|_4y2=  0.  48.   (y  -  a-)2  =  16  -  a-2. 

38.  ^-y2- 2a +  4 y- 4  =  0.  '  49.  (x  +  y)a  =  y2 (y  + 1). 

39.  9a2  +  36y2-96y  +  28  =  0.  50.  a2-4ay  -5y2  +  9  y4  =  0. 

40.  a3  +  3  a;2  -  y2  -  a;  -  3  =  0. 

41.  y3  =  a(a2-9). 

42.  y3  =  a2  (a  4- 3). 

43.  (y  +  l)3=(;*4-l)(a2-9). 

44.  a-2  —if  {%  +  y)  =  0. 

45.  x2  —  if  4-  if  4-  2  y  =  0. 

46.  (y  +  3)2  =  x  (x  -  2)2. 

47.  (y-2)a  =  (x-2)2(a;-5).  55.  y2(a2+  a2)  =  a2(a2 


51. 

Q  +  ^  =  1 

52. 

(DM!)- 

53. 

a;*  4-  y *  =  a* 

54 


■(1) +(!)-• 


PROBLEMS  37 

56.  aY  +  b2x4  =  a2b2x2.  m      .  _       ,     1 

J  67.   4y  =  2»  +  - — 

57.  16aY=Px3(a2—2ax).  2x 

58.  xy  =  16.  68.  y  =  x  +  —  • 

59.  xy  =  —  16.  3 

61.  fr  +  iy.-iy  TO.  ,-«>  +  §. 


^^F+V'^W 


63.  jf  =  -j 


x2  —  5  a;  —  6 
x-1 


72.  xy2  =  4a2(2a-x). 

*3 

J        2  a  —  x 


2  _  X   ~X 

'   *    -  (x  +l)(x  +  3)'  ?4  _x2f„+x) 

65.  x2y2  +  25  =  9/ 


:  BW. 


x(x+l)  75.  y9(a5a  +  a9)  =  a«a 

■'  a  - 1  76.   I/O-'  +  a*)  =  a-(u  -  x). 

Find  the  points  of  intersection  of  the  following  pairs  of  loci : 

77.  3 x  _  j,  _  2  =  0,  5 x  -  3 //  +  2  =  0. 

78.  6 x  -  24 y  +19  =  0,  12'x  +  3y  +  4  =  0. 

79.  2  x  -  y-  2  =  0,  Xs  +  y2  =  25. 

80.  2 x  -  3 y  +  9  =  0,  ./•-  +  y*  +  2x  +  4  y  -8  =  0. 

81.  4  x  +  5  y  -  20  =  0,  x-  +  y2  -  2x  -  3  =  0. 

82.  2y  +  3x-5=  0,  Xs  -  2x  -  2y  +  4  =  0. 

83.  3  x  -  2  y  +  G  =  0,   if  +  4  y  +  x  +  7  =  0. 

84.  x  -  4  y  +  1  =  0,  4  y2  +  4  y  -  4  x  +  5  =  0. 

85.  3  x  +  2  y  -  7  =  0,  5  x2  +  4  y*  =  21. 

86.  7x  -  2  y  +  4  =  0,   21  x9  -  4  //-  -  12  =  0. 

87.  x  -  2  y  =  0,  xV  +  30  =  25  y2. 

88.  2x-y-l=0,  4y2=(x  +  2)(2x-l)2. 

89.  x  +  2  y  -  2  =  0,  y  +  x2//  =  1. 

90.  x2  +  y2  =  25,  16  x2  +  27  y2  =  576. 

91.  Xs  +  y2  =  12,  x2  -  8  y  +  8  =  0. 
4xy=l,  2x2  +  2y2  =  l. 


92 


38  GRAPHS  OF  ALGEBRAIC  FUNCTIONS 

93.  32//2-9:z8  =  0,  %f-§x  =  0. 

94.  2y*  =  3  -x,  y2  =  T7^ — 

95.  x2  -  f  =  0,  x2  +  f  -  4  y  -  4  =  0. 

96.  7^  =  25-5^^-2  =  ^. 

97.  jf-2^,  16(v/-2)  =  (a;-l)3. 

Find   the   real   roots,   accurate   to   two   decimal   places,   of   the 
following   equations  : 

98.  xs  +  2x-  6  =  0.  101.  x*  -  4x3  4-  4  =  0. 

99.  a;3  +  x  + 11  =  0.  102.  x*  -  3  a;a  +  6  x  -  11  =  0. 
100.  a;4  - 11  x  +  5  =  0.                       103.  x3  +  3  ar  4-  4  a;  +  7  =  0. 


CHAPTER  III 


CHANGE  OF  COORDINATE  AXES 


16.  Introduction.  So  far  we  have  dealt  with  the  coordinates 
of  any  point  in  the  plane  on  the  supposition  that  the  axes  of  coor- 
dinates are  fixed,  and  therefore  to  a  given  point  corresponds  one, 
and  only  one,  pair  of  coordinates,  and,  conversely,  to  any  pair  of 
coordinates  corresponds  one,  and  only  one,  point.  But  it  is  some- 
times advantageous  to  change  the  position  of  the  axes,  that  is,  to 
make  a  transformation  of  coordinates,  as  it  is  called.  In  such  a 
case  we  need  to  know  the  relations  between  the  coordinates  of  a 
point  with  respect  to  one  set  of  axes  and  the  coordinates  of  the 
same  point  with  respect  to  a  second  set  of  axes. 

The  equations  expressing  these  relations  are  called  formulas 
of  transformation.  It  must  be  borne  in  mind  that  a  trans- 
formation of  coordinates  never  alters  the  position  of  the  point 
in  the  plane,  the  coordinates  alone  being  changed  because  of 
the  new  standards  of  reference 
adopted. 

17.  Change  of  origin.  In  this 
case  a  new  origin  is  chosen,  but 
the  new  axes  are  respectively 
parallel  to  the  original  axes. 

Let  OX  and  0  Y  (fig.  40)  be  the 
original  axes,  and  O'X'  and  O'Y' 
the  new  axes  intersecting  at  0', 
the  coordinates  of  0'  with  respect 
to  the  original  axes  being  xo  and  yQ. 
Let  P  be  any  point  in  the  plane, 

its  coordinates  being  x  and  y  with  respect  to  OX  and  0  F,  and 
x'  and  y'  with  respect  to  O'X'  and  0'  Y'.  Draw  PMM'  parallel 
to  OY,  intersecting  OX  and  O'X'  at  M  and  M'  respectively. 

39 


1 

r             3 

r 

p 

t 

0 

\M 

0' 

x 

M' 

Fig.  46 


40  CHANGE  OF  COORDINATE  AXES 

Then  OM  =  x,  MP =  y, 

0'M'  =  x',  M'P  =  y', 
NO'=x0,    ON=y0. 
But  OM=  NM'  =  NO'  +  O'M ', 

and  MP  =  MM'  +  M'P  =  ON+  M'P. 

Therefore  x  =  x0  +  x',         y  =  yo  +  y', 

which  are  the  required  formulas  of  transformation. 

Ex.  1.  The  coordinates  of  a  certain  point  are  (3,  —  2).  What  will  be 
the  coordinates  of  this  same  point  with  respect  to  a  new  set  of  axes 
parallel  respectively  to  the  first  set  and  intersecting  at  (1,  —  1)  with 
respect  to  OX  and  OY1 

Here  x0  =  1,  y0  —  —  1,  x  =  3,  and  y  =  —  2.  Therefore  3  =  1  +  x'  and 
—  2  =  —  1  +  y',  whence  x'  —  2  and  y'  =  —  1. 

Ex.  2.  Transform  the  equation  y1  —  2  y  —  3  x  —  5  =  0  to  a  new  set  of 
axes  parallel  respectively  to  the  original  axes  and  intersecting  at  the 
point  (-2,  1). 

The  formulas  of  transformation  are  x  =  —  2  +  x',  y  =  1  +  y'.  Therefore 
the  equation  becomes 

(1  +  y'f  -  2  (1  +  30  -  3  (-  2  +  O  -  5  =  0, 

or  y'2  -  3  x'  =  0. 

As  no  point  of  the  curve  has  been  moved  in  the  plane  by  this  transformation, 
the  curve  has  been  changed  in  no  way  whatever.  Its  equation  is  different  because 
it  is  referred  to  new  axes. 

After  the  work  of  transformation  has  been  completed  the  primes 
may  be  dropped.  Accordingly  the  equation  of  this  example  may  be 
written  y2  —  3  x  —  0,  or  y2  =  3  x,  the  new  axes  being  now  the  only  ones 
considered. 

18.  One  important  use  of  transformation  of  coordinates  is 
the  simplification  of  the  equation  of  a  curve.  In  Ex.  2  of 
the  last  article,  for  example,  the  new  equation  y1  =  3  x  is 
simpler  than  the  original  equation.  It  is  obvious,  however, 
that  the  position  of  the  new  origin  is  of  fundamental  im- 
portance in  thus  simplifying  the  equation,  and  we  shall  now 
solve  examples  illustrating  methods  of  determining  the  new 
origin  to  advantage. 


CHANGE  OF  ORIGIN  41 

Ex.  1.  Transform  the  equation  y2  -  4  y  -  x3  -  3  x2  —  3  x  +  3  =  0  to 
new  axes  parallel  respectively  to  the  original  axes,  so  choosing  the 
origin  that  there  shall  be  no  terms  of  the  first  degree  in  x  and  y  in 
the  new  equation. 

The  formulas  of  transformation  are 

x  =  x0  +  x'     and     y  =  y0  +  y', 

where  suitable  values  of  x0  and  y0  are  to  be  determined.  The  equa- 
tion becomes 

(i/o  +  y'f  -  4  0/o  +  y')  -  (*o  +  x'Y  -  3  (*0  +  *y  -  3  (*o  +  «0  +  3  =  °> 

or,  after  expanding  and  collecting  like  terms, 

y*  +  (2^-4)/-  x'3-(3x0  +  3)x'2-(3x2  +  C/0  +  3)x' 

+  G/o2  ~  4  Z/o  -  *o8  ~  3  xo  -  3  X0  +  3)  =  0. 

By  the  conditions  of  the  problem  we  are  to  choose  x0  and  ?/„  so  that 

2#0-4  =  0,         3x^  +  0x0+3  =  0, 

two  equations  from  which  we  find  x0  =  —  1  and  y0  =  2. 

Therefore  (—  1,  2)  should  be  chosen  as  the  new  origin,  and  the  new 
equation  is  y'-  —  x'3  =  0,  or  y"  —  x3  after  the  primes  are  dropped. 

Ex.  2.    Transform  the  equation 

10  x2  +  25  if  +  04  x  -  150  y  -  111  =  0 

to  new  axes  parallel  respectively  to  the  original  axes,  so  choosing  tin- 
origin  that  there  shall  be  no  terms  of  the  first  degree  in  x  and  y  in 
the  new  equation. 

We  may  solve  this  example  by  the  method  used  in  solving  Ex.  1,  but 
since  the  equation  is  of  the  second  degree,  the  following  method  is  very  , 
desirable. 

Rewriting,  we  have 

16  (x2  +  4  x)  +  25  O2  -  6  y)  =  111 ; 
whence  16  (x2  +  4  x  +  4)  +  25  (y*  -  6  y  +  9)  =  400, 
or  10(x  +  2)'-+25(?/-3)2=400. 

Placing  now  x  =  —  2  +  x/,     y  =  3  +  y', 

we  have  as  our  new  equation         16  x'2  +  25  y'2  =  400, 

the  new  origin  of  coordinates  being  at  the  point  (—  2,  3)  with  respect  to 
the  original  axes. 


42 


CHANGE  OF  COORDINATE  AXES 


19.  Change  of  direction  of  axes. 

Case  I.  Rotation  of  axes.  Let  OX  and  OY  (fig.  47)  be  the 
original  axes,  and  OX'  and  OY'  be  the  new  axes,  making  Z  $ 
with  OX  and  OY  respectively.  Then  Z XOY' =  90°  +  <f>  and 
ZYOX'=  90°-  (f>. 

Let  P  be  any  point  in  the  plane, 
its  coordinates  being  x  and  y  with 
respect  to  OX  and  OY,  and  a/ 
and  y  with  respect  to  OX'  and 
OY'.  Then,  by  construction,  OM=xy 
OX=y,  OM'  =  x',  and  M'P  =  y'. 
Draw  OP. 

The  projection  of  OP  on  OX  is 
OJf,     and     the     projection     of     the 

broken  line  OM'P  on  OX  is  OM'  cos  <£  +  Jf  'P  cos  (90°  +  <£),  or 
OJf'  cos  </>  -  M'P  sin  </>. 

Therefore  0 M  =  OM'  cos  (/>  -  M'P  sin  0, 

by  §  2. 

In  like  manner  the  projection  of  OP  on  OY  is  (9  A7",  and  the 
projection  of  the  broken  line  OM'P  on  OY  is  OJf'cos  (90°-  <£) 
+  ilPPcos<£. 

Therefore  OA7"  =  OM'  sin  <£  +  M'P  cos  (/>, 

by  §  2. 

Replacing  OJf,  0A",  OJf',  M'P  by  their  values,  we  have 

x  =  x'  cos  (f>  —  y'  sin  0, 
y  =  x1  sin  <j>  +  y'  cos  </>. 

Ex.  1.    Transform  the  equation  xy  =  5  to  new  axes  having  the  same 
origin  as  the  original  axes  and  making  an  angle  of  45°  with  them.  ,         , 
Here   <j>  =  45°,  and  the  formulas  of  transformation  are  x  = 

V        V2 

Substituting  and  simplifying,  we  have  as  the  new  equation  x2  —  y1  =  10. 

Ex.  2.  Transform  the  equation  34  x1  +  41  y2  —  24  xy  =  100  to  new  axes 
with  the  same  origin  as  the  original  axes,  so  choosing  the  angle  <£  that  the 
new  equation  shall  have  no  term  in  xy. 


V2 


OBLIQUE  COORDINATES  43 

The  formulas  of  transformation  are 

x  =  x'  cos  <f>  —  y'  sin  <£, 
y  =  x'  sin  <j>  +  y  cos  </>, 

where  <f>  is  to  be  determined. 

Substituting  in  the  equation  and  collecting  like  terms,  we  have 

(34  cosa<£  +  41  sin2<f>  -  24  sin  <£  cos  (f>)  x2 

+  (34  sin2  <f>  +  41  cos2  <f>  +  24  sin  <£  cos  <£)  y» 

+  (24  sin2<£  +  14  sin  <f>cos<f>-  24  cos2<£)  xy  =  100. 

By  the  conditions  of  the  problem  we  are  to  choose  <f>  so  that 
24  sin2</>  +  14  sin<£  cos  </>  —  24  cos2<f>  =  0. 

One  value  of  <f>  satisfying  this  equation  is  tan-1 1.  Accordingly  we  sub- 
stitute sin  <£  =  i?-  and  cos  <f>  =  $,  and  the  equation  reduces  to  x1  +  2  y1  =  4. 

Case  II.  Interchange  of  axes.  If  the  axes  of  x  and  y  are  simply 
interchanged,  their  directions  are  changed,  and  hence  such  ii  trans- 
formation is  of  the  type  under  consideration  in  this  article.  Hie 
formulas  for  such  a  transformation  are  evidently  x  —  y',y  =  x'. 

Case  III.  Rotation  and  interchange  of  axes.  Finally,  if  the 
axes  are  rotated  through  an  angle  <f>  and  then  interchanged,  the 
formulas,  being  merely  a  combination  of  the  two  already  found,  are 

x  =  y'  cos  <£  —  x'  sin  <f>,  y  =  y'  sin  </>  +  x'  cos  <f>. 

A  special  case  of  some  importance  occurs  when  <f>  =  270°.  We 
have  then  x  =  x',  y  =—  y'. 

Cases  II  and  III,  it  should  be  added,  occur  much  less  frequently 
than  Case  I. 

If  both  the  origin  and  the  direction  of  the  axes  are  to  be 
changed,  the  processes  may  evidently  be  performed  successively, 
preferably  in  this  order :  (1)  change  of  origin ;  (2)  change  of 
direction. 

20.  Oblique  coordinates.  Up  to  the  present  time  we  have 
always  constructed  the  coordinate  axes  at  right  angles  to  each 
other.  This  is  not  necessary,  however,  and  in  some  problems, 
indeed,  it  is  of  advantage  to  make  the  axes  intersect  at  some 
other  angle.  Accordingly,  in  fig.  48,  let  OX  and  OY  intersect 
at  some  angle  to  other  than  90°. 


44 


CHANGE  OF  COORDINATE  AXES 


We  now  define  x  for  any  point  in  the  plane  as  the  distance 
from  OF  to  the  point,  measured  parallel  to  OX,  and  y  as  the 
distance  from  OX  to  the  point,  measured  parallel  to  OY.    The 
algebraic  signs  are  determined  accord- 
ing to  the  rules  adopted  in  §  4. 

It  is  immediately  evident  that  the 
rectangular  coordinates  are  but  a  special 
case  of  this  new  type  of  coordinates, 
called  oblique  coordinates,  since  the  new 
definitions  of  x  and  y  include  those  pre- 
viously given.  In  fact,  the  term  Carte- 
sian, or  rectilinear,  coordinates  includes 
both  the  rectangular  and  the  oblique. 

Oblique  coordinates  are  usually  less  convenient  than  rectangu- 
lar ones  and  are  very  little  used  in  this  book.  If  necessary,  the 
formulas  obtained  by  using  rectangular  coordinates  can  be  trans- 
formed into  similar  ones  in  oblique  coordinates  by  the  formulas 
of  the  following  article.  When  no  angle  is  specified  the  angle 
between  the  axes  is  understood  to  be  a  right  angle. 

21.  Change  from  rectangular  to  oblique  axes.  Let  OX  and  OY 
(fig.  49)  be  the  original  axes,  at  right  angles  to  each  other,  and 
OX'  and  OY'  the    new  Y 

axes,  making  angles  (f> 
and  (f>'  respectively  with 
OX.  Then  co  =  $  -  <j>. 
Let  P  be  any  point  in 
the  plane,  its  rectangular 
coordinates  being  x  and 
y,  and  its  oblique  coordi- 
nates being  x'  and  y'. 
Draw  PM  parallel  to  OY,  PM'  parallel  to  OY',  M'N  parallel 
to  OY,  and  RM'N'  parallel  to  OX.     Then  ZBM'P=<f>'. 

But  031=  ON  +  NM  =  ON  +  M'N'  =  031'  cos  <f>  +  31' P  cos  <f>', 
3IP  =  3IN'+  N'P  =  N3f  +  N'P  =  031'  sin  (f>  +  3f'P  sin  </>'. 
Therefore  x  —  x'  cos  <$>  +  y'  cos  </>', 

y  =  x'  sin  <j>  +  y'  sin  <f>'. 


p 

/ 

^-x' 

'-^v 

-R 

^V\       ! 

\             M 

N 

Fig.  49 

PROBLEMS  45 

Ex.    Transform  the  equation  — -  —  "—  =  1  to  the  lines  y  =  ±  -  x  as  axes. 
«-       b-  a 

Here  let  <f>  =  tan-1( — '-),  and  <£'  =  tan-1--     The  formulas  of  trans- 
_..       , \     a]  a 


formation  become 


0'+  y'),      y  =   ,- — -  (-  x'  +  /)• 


Vo2  +  b-  Va2  +  //- 

a2  +  b- 

Substituting  and  simplifying,  we  have  as  the  new  equation  xy  =  — 

i  4 

Unless  b  =  a,  the  axes  are  oblique  and  w  =  2  tan-1-- 

a 

22.  Degree  of  the  transformed  equation.  In  reviewing  this 
chapter  we  see  that  the  expressions  for  the  original  coordinates  in 
terms  of  the  new  are  all  of  the  first  degree.  Hence  the  result  of 
any  transformation  cannot  be  of  higher  degree  than  the  original 
equation.  On  the  other  hand,  the  result  cannot  be  of  lower 
degree  than  the  original  equation  ;  for  it  is  evident  that  if  any 
equation  is  transformed  to  new  axes  and  then  back  to  the  original 
axes,  it  must  resume  its  original  form  exactly.  Hence,  it'  the 
degree  had  been  lowered  by  the  first  transformation,  it  must  be 
increased  to  its  original  value  by  the  second  transformation. 
But  this  is  impossible,  as  we  have  just    noted. 

It  follows  that  the  degree  of  an  equation  is  unchanged  by 
any  single  transformation  of  coordinates  or  by  any  number  of 
successive  transformations.  In  particular,  the  proposition  that 
any  equation  of  the  first  degree  represents  a  straight  line  is 
true  for  oblique,  as  for  rectangular,  coordinates. 

PROBLEMS 

1.  What  are  the  new  coordinates  of  the  points  (3,  4),  (—  3,  6), 
and  (4,  —  7)  if  the  origin  is  transferred  to  the  point  (2,  —  3),  the 
new  axes  being  parallel  to  the  old  ? 

2.  Transform  the  equation  a-2  +  9//2  —  4.z+18?/  +  8  =  0to  new 
axes  parallel  to  the  old  axes  and  meeting  at  the  point  (2,  —  1). 

3.  Transform  the  equation  2x2  +  2?/2  —  2x  -{-  2  >/  —  7=0  to  new 
axes  parallel  to  the  old  axes  and  meeting  at  the  point  (^,  —  £). 

4.  Transform  the  equation  a-2  —  if  +  2x  —  3  =  0  to  new  axes 
parallel  to  the  old  axes  and  meeting  at  the  point  (—  1,  0). 


46  CHANGE  OF  COOBDINATE  AXES 

5.  Transform  the  equation  f  -  3  if  +  3  x2  +  3y/-fl2*+ll=0 
to  new  axes  parallel  to  the  old  axes  and  meeting  at  the  point  (—  2,  1). 

6.  Transform  the  equation  f  —  4  a*  —  6//  +  5  =  0  to  new  axes 
parallel  to  the  old  axes,  so  choosing  the  new  origin  that  the  new 
equation  shall  contain  only  terms  in  if  and  x. 

7.  Transform  the  equation  x2  +  2x  +  4?/  —  3  =  0  to  new  axes 
parallel  to  the  old,  so  choosing  the  new  origin  that  the  new  equation 
shall  contain  only  terms  in  x2  and  y. 

8.  Transform  the  equation  2  x2  —  4  //  + 12  x  + 16  y  —  7  =  0  to 
new  axes  parallel  to  the  old,  so  choosing  the  origin  that  there  shall 
be  no  terms  of  the  first  degree  in  the  new  equation. 

9.  Transform  the  equation  4a;2  +  9  y2  —  Ax  +12 y  +  4  =  0  to 
new  axes  parallel  to  the  old,  so  choosing  the  origin  that  there 
shall  be  no  terms  of  the  first  degree  in  the  new  equation. 

10.  Transform  the  equation  xy  —  3  y  +  2  x  — 12  =  0  to  new  axes 
parallel  to  the  old,  so  choosing  the  origin  that  there  shall  be  no  terms 
of  the  first  degree  in  the  new  equation. 

11.  Transform  the  equation  6xy  —  10.x  +  3  y  — 19  =  0  to  new 
axes  parallel  to  the  old,  so  choosing  the  origin  that  there  shall  be 
no  terms  of  the  first  degree  in  the  new  equation. 

12.  Show  that  any  equation  of  the  form  xy  +  ax  -4-  by  -f-  c  =  0 
can  always  be  reduced  to  the  form  xy  =  k  by  choosing  new  axes 
parallel  to  the  old,  and  determine  the  value  of  k. 

13.  Show  that  the  equation  y2  -\-  ay  +  hx  -f-  c  =  0  (b  ^=  0)  can  always 
be  reduced  to  the  form  if  +  bx  =  0  by  choosing  new  axes  parallel  to 
the  given  ones. 

14 .  Show  that  the  equation  ax2 + by2  -f-  ex  -\-  dy  -\-e  =  0(a=f=0,b=£0) 
can  always  be  put  in  the  form  ax2  +  by2  =  k  by  choosing  new  axes 
parallel  to  the  old,  and  determine  the  value  of  k. 

15.  What  are  the  coordinates  of  the  points  (0,  2),  (2,  0),  (2,  -  2) 
if  the  axes  are  rotated  through  an  angle  of  60°? 

16.  What  are  the  coordinates  of  the  points  (1,  2),  (2,  2),  (2,  -1) 
if  the  axes  are  rotated  through  an  angle  of  45°? 

17.  What  are  the  coordinates  of  the  points  (1,  2),  (—1,  —2), 
(1,  —  2)  if  the  axes  are  rotated  through  an  acute  angle  tan-1!? 

18.  Transform  the  equation  2x2  -f  2  y2  —  3xy  —  7=  0  to  a  new 
set  of  axes  by  rotating  the  original  axes  through  an  angle  of  45°. 


PROBLEMS  47 

19.  Transform  the  equation  4.r2  +  2  ~v3xy  +  2  f  —  5  =  0  to  a 
new  set  of  axes  by  rotating  the  original  axes  through  a  positive 
angle  of  30°. 

20.  Transform  the  equation  4  .r2  — 12  xy  +  9  f  —  14  =  0  to  a  new 
set  of  axes  making  a  positive  angle  tan-1f  with  the  original  set. 

21.  Transform  the  equation  5  .r2  -f  4  xy  +  8  f  —  36  =  0  to  a  new 
set  of  axes  by  rotating  the  original  axes  through  a  positive  angle 
tan-1(—  i). 

22.  Transform  the  equation  4  x2,  + 15  xy  —  4  if  —  34  =  0  to  a  new 
set  of  axes  making  a  positive  angle  tan-1|  with  the  original  axes. 

23.  Show  that  the  equation  x*  +  f  =  <r  will  be  unchanged  in 
form  by  transformation  to  any  pair  of  rectangular  axes  if  the  origin 
is  unchanged. 

24.  Transform  the  equation  x*  —  f  =  49  to  new  axes  bisecting 
the  angles  between  the  original  axes. 

25.  Transform  the  equation  5 .,■-  +  2 xy  +  5  if-  — 12  =  0  to  one 
which  has  no  cry-term,  by  rotating  the  axes  through  the  proper 
angle. 

26.  Transform  the  equation  C>  ./■- -f- 2  I ./// —  y2  —  150  =  0  to  one 
which  has  no  xy-term,  by  rotating  the  axes  through  the  proper 
angle. 

27.  Transform  the  equal  ion  1 6  x*  -  2  I  xy  +  9  //"  -  30  x  -  40  y  =  0 
to  one  which  has  no  scy-tenn,  by  rotal  ing  the  axes  through  the  proper 
angle. 

28.  Transform  the  equation  11  x*  -f  2  1  .#•//  +  34  f  -100. r  -  50// 
—  100  =  0  to  one  which  has  no  xy-tevm.,  by  rotating  the  axes  through 
the  proper  angle. 

29.  Transform  the  _  equation  1 1  .r-  -  (')  V:'»  xy  +  5  f  +  (22  — 
12  V3)x  -(20  +  G  V3)y  +3-12  V3  =  0  to  a  new  set  of  axes 
making  an  angle  of  G0°  with  the  original  axes  and  intersecting  at 
the  point  (—1,  2)  with  respect  to  the  original  axes. 

30.  Transform  the  equation  4  a-2  +  25  f  =  100  from  rectangular 
axes  to  oblique  axes  with  the  same  origin  and  making  the  angles 
tan-1  J  and  tan_1(— f)  respectively  with  OX. 

31.  Transform  the  equation  9  x2  —  4  f  =  36  from  rectangular  axes 
to  oblique  axes  with  the  same  origin  and  making  the  angles  tan-1| 
and  tan-1(—  |)  respectively  with  OX. 


48  CHANGE  OF  COORDINATE  AXES 

32.  Transform  the  equation  9  x2  —  4  if  =  36  from  rectangular  axes 
to  oblique  axes  with  the  same  origin  and  making  the  angles  tan_1| 
and  tan-1 3  respectively  with  OX. 

33.  Prove  that  the  formulas  for  changing  from  a  set  of  rectangular 
axes  to  a  set  of  oblique  axes  having  the  same  origin  and  the  same 
axis  of  x  are  x  =  x' +  y' cos  u>, 

y  =  y'  sin  w, 
where  <o  is  the  angle  between  the  oblique  axes. 

34.  By  rotating  the  axes  through  an  angle  of  45°  and  changing 
the  origin,  prove  that  the  equation  x1  +  y-  =  a?  can  be  transformed 
into  y2  =  V2  ax,  and  sketch  the  curve. 

35.  The  equation  of  the  Folium  of  Descartes  is  xB  +  ys  —  3  axy  =  0. 
Rotate  the  axes  through  an  angle  of  45°  and  sketch  the  curve. 


CHAPTER  IV 


GRAPHS  OF  TRANSCENDENTAL  FUNCTIONS 


23.  Definition.  Any  function  of  x  which  is  not  algebraic  is  called 
transcendental.  The  elementary  transcendental  functions  are  the 
trigonometric,  the  inverse  trigonometric,  the  exponential,  and  the 
logarithmic  functions,  the  definitions  and  the  simplest  properties 
of  which  are .  supposed  to  be  known  to  the  student.  In  this 
chapter  we  shall  discuss  the  graphs  of  these  functions. 

24.  Trigonometric  functions. 

Ex.  1.   y  =  sinx. 

The  values  of  y  are  found  from  a  table  of   trigonometric   functions. 
In  plotting  it  is  desirable  to  express  x  in  circular  measure;   for  example, 
for  the  angle  180°  we  lay  off 
x  =  it  =  :$.141l).     When  x  is  a 
multiple   of   it,   y  —  0  ;    when 

x  is  an  odd  multiple  of  -, 

y  =  ±  1 ;  for  other  values  of  x, 

y  is  numerically  less  than  1. 

The    graph    consists    of    an 

indefinite  number  of  congruent  arches,  alternately  above  and  below  the 

axis  of  x,  the  width  of  each  arch  being  it  and  the  height  being  1  (fig.  50). 


The  curve  y  =  sin  x  may  be  constructed  without  the  use  of  tables,  by  a 
method  illustrated  in  fig.  51. 

Let  Pt  be  any  point  on  the  circumference  of  a  circle  of  radius  1 
with  its  center  at  C,  and  let  AO  be  a  diameter  of  the  circle  extended 

40 


50   GKAPHS  OF  TRANSCENDENTAL  FUNCTIONS 

indefinitely.  With  a  pair  of  dividers  lay  off  on  AO  produced  a  distance 
ONx  equal  to  the  arc  OPv  This  may  be  done  by  considering  the  arc  OP1 
as  composed  of  a  number  of  straight  lines  each  of  which  differs  inappre- 
ciably from  its  arc.  From  JVj  draw  a  line  perpendicular  to  AO,  and  from 
Px  draw  a  line  parallel  to  AO.  Let  these  lines  intersect  in  Qv  Then 
N1Ql  =  M1P1  =  CP1  sin  OCPv  But  CP1  =  1,  and  the  circular  measure 
of  OCPl  is  OP1  =  ONv  If,  then,  we  take  ONx  =  x,  N1Q1  =  y,  Q1  is  a 
point  of  the  curve  ?/  =  sin:r.  By  varying  the  position  of  the  point  Pl 
we  may  construct  as  many  points  of  the  curve  as  we  wish.  The  figure 
shows  the  construction  of  another  point  Qr 

Ex.  2.    y  —  a  sin  bx. 

When  a:  is  a  multiple  of  -i  y  =  0;  when  x  is  an  odd  multiple  of  — -, 
b    '  2  b 

y  =  ±  a ;    for  all  other  values  of  x,  y  is  numerically  less  than  a.    The 

Y 


-SW     1 

-37T  \ 

-7T 

7T 

1      3TT 

/  i^ 

77T 

T  \ 

1      \ 

I0 

\       i 

T 

-,'7TJ 

-37T 
2 

-7T/ 

Ltt\ 

\-' 

I71* 

i.v7T 

»5T 

Pig.  52 


curve  is  similar  in  its  general  shape  to  that  of  Ex.  1,  but  the  width 
of  each  arch  is  now  -?  and  its  height  is  a.  Fig.  52  shows  the  curve 
when  a  =  3  and  b  —  2. 

Ex.  3.    y  =  a  sin  (bx  +  c). 

Place  x  — V  x',  y  =  y'. 

o 
The  equation  then  becomes  y'  =  a  sin  bx'. 

The  graph  is  consequently  the  same  as  in  Ex.  2,  the  effect  of  the  term 
+  c  being  merely  to  shift  the  origin. 


Ex.  4.    y  =  a  cos  bx. 
This  may  be  written 


y  =  a  sin  ( bx  + 


which  is  a  curve  of  Ex.  3.    Hence  the  graph  of  the  cosine  function  differs 
from  that  of  the  sine  function  only  in  its  position. 


TRIGONOMETRIC  FUNCTIONS 


51 


Ex.  5.    y  =  sin  x  +  |  sin  2  x 

The  graph  is 
found  by  adding 
the  ordinates  of 
the  two  curves 
y  =  sin  x      and 

y  - 


4-  sin  2x 


sin  x  +  -jsin  2x 
y=sinx 


1 
l-K,  that  is,  when  x  —  —  i  where  A-  is  any  integer.    Hence 


sin  2  x,  as 
shown  in  fig.  53. 

Ex.  6.    y  =  si 

w  =  0  when  - 

the  graph  crosses  the  axis  of  x  at  the  points  1,  h,  \,  \,  £,  etc.  Between  any 
consecutive  two  of  these  points  y  varies  continuously  from  0  to  ±  1  and  back 
to  zero.  It  follows  that  as 
X  approaches  0,  the  corre- 
sponding point  on  the 
graph  oscillates  an  infi- 
nite number  of  times 
back  and  forth  between 
the  straight  lines  y  =  ±1. 
It  is  therefore  physically 
impossible  to  construct 
the  graph  in  the  neigh- 
borhood of  the  origin. 
This  is  shown  in  fig.  51  Fig.  54 

by  the  break  in  the  curve. 

The  value  of  y  can  be  calculated  for  any  value  of  x,  no  matter  how 

small.     For   example,  if  x  -    —  ,  y  =  sin —  =  .9G59.    The  value  of  y  is 

not  defined  for  x  =  0,  and  the  function  is  discontinuous  at  that  point. 

Ex.  7.    y  =  tan  x. 

When  x  is  a  multiple  of 
7r,  y  =  0 ;  when  x  is  an  odd 
multiple  of  — »  y  is  infinite, 

in  the  sense  of  §  13.  The 
curve  has  therefore  an  un- 
limited number  of  asymp- 
totes perpendicular  to  OX, 


i         i:         ii 

i 

i           ' 

i           ! 

/'         /' 
/!        /! 

/  !     /  I 

' 

O    T\       /|T   3£       /J IT    69 

2,  /  j     a\  /        3 

Fig.  55 


.  • .,  which  divide  the  plane 
into  an  infinite  number  of 
sections,  in  each  of  which  is  a  distinct  branch  of  the  curve,  as  shown  in  fig.  55. 


52        GRAPHS  OF  TRANSCENDENTAL  FUNCTIONS 


25.  Inverse  trigonometric  functions.  The  graphs  of  the  inverse 
trigonometric  functions  are  evidently  the  same  as  those  of  the 
direct  functions  but  differently  placed  with  reference  to  the 
coordinate  axes.  It  is  to  be  noticed  particularly  that  to  any 
value  of  x  corresponds  an  infinite  number  of  values  of  y. 


Ex.  l. 


From  this,  x  =  sin  y,  and  we  may  plot  the  graph 
by  assuming  values  of  y  and  computing  those  of  x 
(fig-  56). 


Fig.  56 


Fig.  57 


Ex.  2.    y  =  tan-1:*;. 

Then  x  =  tan  y,  and  the  graph  is  as  in  fig.  57. 

These  curves  show  clearly  that  to  any  value  of  x  corresponds  an  infinite 
number  of  values  of  y. 

26.  Exponential  and  logarithmic  functions.    The  equation 

y  =  of 

defines  y  as  a  continuous  function  of  x,  called  the  exponential  func- 
tion, such  that  to  any  real  value  of  x  corresponds  one  and  only 
one  real  positive  value  of  y.  A  proof  of  this  statement  depends 
upon  higher  mathematics,  but  the  student  is  already  familiar  with 
the  methods  by  which  the  value  of  y  may  be  computed  for  simple 
values  of  x.    If  x  is  an  integer  n,  y  is  determined  by  raising  a  to 

the  nth.  power  by  multiplication.    If  a;  is  a  positive  fraction  !- ,  y  is 

the  qth.  root  of  the  pth  power  of  a.  If  x  is  a  positive  irrational  num- 
ber, the  approximate  value  of  y  may  be  obtained  by  expressing  x 


EXPONENTIAL  AND  LOGARITHMIC  FUNCTIONS     53 


approximately  as  a  rational  number.   If  x  =  0,  y  =  a0  =  1.    Finally, 
if  x  —  —  ?n,  where  m  is  any  positive  number,  y  =  a~m  =  —  • 

Practically,  however,  the  value  of  ax  is  most  readily  obtained 
by  means  of  the  inverse  function,  the  logarithm;  for  if 

y  =  ax, 
then  x  =  loga  y. 

The  quantity  a  is  called  the  base  of  the  system  of  logarithms 
and  may  be  any  number  except  1. 

When  a  —  10,  tables  of  logarithms  are  readily  accessible.  Sup- 
pose a  is  not  10,  and  let  b  be  such  a  number  that 

10"=  a; 
6  =  log10a. 

y  =  ax  =  (10")x  =  106*. 


that  is, 

Then  we  have 
Hence 


and 


_  logoff  _  logi 


b  log10« 

Ex.  1.    The  graph  of  y  =  log(li)x  is  shown  in  fig.  58. 
It  is  to  be  noticed  that  the  curve  has  the  negative  portion  of  the  //-axis 
for  an  asymptote  and  has  no  points  corresponding  to  negative  values  of  x. 

Ex.  2.  The  graph  of  y  =  (1.5)*  is  shown  in  fig.  59. 


Fig.  58 

27.  The  number  e.  In  the  theory  and  the  use  of  the  expo- 
nential and  the  logarithmic  functions  an  important  part  is 
played  .by  a  certain  irrational  number,  commonly  denoted  by 
the  letter  e.   This  number  is  denned  by  an  infinite  series,  thus: 


54   GRAPHS  OF  TRANSCENDENTAL  FUNCTIONS 

It  can  be  shown  that  this  series  converges ;  that  is,  that  the 
greater  the  number  of  terms  taken  the  more  nearly  does  their 
sum  approach  a  certain  number  as  a  limit.  Assuming  this,  we 
may  compute  e  to  seven  decimal  places  by  taking  the  first 
eleven  terms.  There  results 

e  =  2.7182818.... 


When  y  =  ex,  x  is  called  the  natural,  or  Napierian,  logarithm 
of  y.  The  use  of  Napierian  logarithms  in  theoretical  work  gives 
simpler  formulas  than  would  result  from  the  use  of  the  common 
logarithm.  Hence  in  theoretical  discussions  the  expression  log  x 
usually  means  the  Napierian  logarithm.  On  the  other  hand, 
when  the  chief  interest  is  in  calculation  of  numerical  values,  as 
in  the  solution  of  triangles,  \ogx  usually  means  log10#.  In  this 
book  we  shall  use  log x  for  logex. 

Tables  of  values  of  logea:  and  ex  are  found  in  many  collections 
of  tables  and  may  be  used  in  finding  the  graphs.    It  is  evident, 
however,   that   the   graphs 
will   not  differ  in  general 
shape  from  those  in  Exs.  1 
and  2  of  §  26. 

In  the  following  exam- 
ples we  give  the  graphs 
of  certain  other  functions 
which  involve  e  and  pre- 
sent other  points  of  interest. 

Ex.  1.  y  =  er< 

The  curve  (fig.  60)  is  symmetrical  with  respect 
to  OY  and  is  always  above  OX.  When  x  =  0,  y  =  1. 
As  x  increases  numerically,  y  decreases,  approaching 
zero.    Hence  OX  is  an  asymptote. 


Fig.  60 


Ex. 


:(e"  +  e    «). 


This  is  the  curve  (fig.  61)  made  by  a  string 
held  at  the  ends  and  allowed  to  hang  freely. 
It  is  called  the  catenary. 


Fig.  61 


PROBLEMS 


55 


Ex.  3.   y  =  e-^smbx. 

The  values  of  y  may  be  computed  by  multiplying  the  ordinates  of 
the  curve  y  =  e-"*  by  the  values  of  sin  bx  for  the  corresponding 
abscissas.  Since  the  value  of 
sin  bx  oscillates  between  1  and  —  1, 
the  values  of  e~  "*  sin  bx  cannot 
exceed  those  of  e_ar.  Hence  the 
graph  lies  in  the  portion  of  the 
plane  between  the  curves  y  =  e~  ■" 
and  y  =  —  e-ax.  When  x  is  a  multi- 
ple of  — ,  y  is  zero.  The  graph  there- 
o 

fore  crosses  the  axis  of  x  an  infinite 
number  of  times.  Fig.  G2  shows 
the  graph  when  a  =  1,  b  =  2  ir. 


Ex.  4.   y  =  ex. 

When  x  approaches   zero,  being  positive, 
When  x  approaches  zero,  being   negative, 
y    approaches    zero ;     for    example,    when 

"  T  o\  (T> 

The  function  is  therefore 


p-1000  —  . 


increases   without   limit. 
7 


=  e1000,  and  when 

*  e1000 

discontinuous  for  x  =  0. 

The  line  y  =  1  is  an  asymptote  (fig.  63),  for 
as  x  increases  without  limit,  being  positive  or 

negative,  -  approaches  0,  and  y  approaches  1. 

1  +  (i* 
As    x    approaches    zero    positively,    y    ap- 
proaches  zero.    As  x  approaches  zero   nega- 
tively,   y    approaches    10.      As    x    increases 
indefinitely,   y  approaches   5. 

The  curve  (fig.  6-i)  is  discontinuous  when 
x  =  0. 

PROBLEMS 
Plot  the  graphs  of  the  following  equations 


Fig. 


Pig.  64 


1.  y  =  £  sin  2x. 

2.  2/  =  3sin^- 


4.  y  =  sin(  x 


5.  y  =  2s'm3(x  —  --)• 


3.  y=s8m[x  + 


6.  y 


x-1 


50   GRAPHS  OF  TRANSCENDENTAL  FUNCTIONS 


7.  y  =  isin(2aj  +  3).  /        *■> 

30.  y 

I.  y  =  cos  3x. 

).  y  =  3  cos  -j-  31.  y 


17.  y  =  sin  -—  —  -  sin  7ra: 


>(<*-!) 


30.   ?/  =  sec  (a; 
8.  y  =  cos3 

i-  31.  y  =  sm  *  — 

10.  y  =  2  cos  3  (a;  +  2).  32-  2/  =  cos^x  +  2). 

li.  y  =  2  cos  (2*-l).  33>  y^^-i^Z". 

12.  y=  versa;.  *  "+"  x 

13.  y-2  +  BnSft  34'  y  =  ta^>+1)- 

14.  y  =  2  —  icosa;.  35.  y  =  tan"1-— — -• 

15.  y  =  sin  x  -f  sin  3  a;.  36.  y  =  ex~x. 

16.  y  =  \  sin  a*  —  \  sin  2a\  37.  y  =  xe~x. 


irx       1    .  38.  y  =  a^e-*. 

i 
39.   y  =  ice*. 


41.  y  =  e1-*. 

l  +  x 


19.  y  =  1  +  cos  x  —  £  cos  3  x.  40.  y  =  xe  x 

20.  y  =  sin2a;. 

21.  y  =  sin  a;2. 

1  42.   y  =  e1^*. 

22.  y  =  a;  sin-« 

a;  43.  y  =  \{f  —  e~x). 

23.  y  =  a;2sin-.  44.  y  =  £(e*  +  e"*). 

24.  y  =  etna;.  45.  y 


25.  y  =  i  tan  2a. 

26.  y  =  2tan-«  47.  y  =  e~Sx  sin  2jb. 


46.  y  =  e-*  eosar. 


2 


27.  y  =  4tan^p.  48.  y  =  log  ~ 

28.  y  =  seca\  49.  y  =  log  sin  x. 

29.  y  =  csca;.  50.  y  =  log  tana;. 


CHAPTER  V 

THE  STRAIGHT  LINE 

28.  The  point-slope  equation.  If  the  slope  of  a  straight  line 
and  a  point  on  the  line  are  known,  the  equation  of  the  line  is 
readily  found.  Let  LK  (fig.  65)  be  any  straight  line,  I^(xvy^) 
a  known  point  on  it,  and  m  its  slope.  TakeP(x,  y~),  any  point 
on  the  line.    Then,  by  §  6, 

lzh=m. 

x  —  x 

If  m  is  not  infinite,  we  may  clear 
of  fractions  and  obtain 


y-yx  =  m(x-xj. 


Fig.  65 


This    is    an    equation    which 
obviously   satisfied    by   the    coordi- 
nates  of   any  point   on  LK  and   by  those   of  no   other   point. 
Hence  it  is  the  equation  of  LK. 

If  the  line  is  parallel  to  OX,  m  —  0,  and  the  equation  of  the 
line  is  ,ON 

y=yx-  (2) 

If  the  line  is  parallel  to  0  Y,  m  =  <x,  and  the  equation  of  the 

line  is  .Q\ 

x  =  xx.  (3) 

Ex.    Find  the  equation  of  a  straight  line  with  the  slope  —  \,  passing 
through  the  point  (5,  7). 

By  substituting  in  the  formula,  we  have 


whence 


y-7 

2x  +  3^-31 


10-5); 


29.  The  slope-intercept  equation.    The  equation  (1)  of  §  28 
takes  a  special  form  when  the  point  L\  is  taken  at  B  (fig.  65), 

57 


58  THE  STRAIGHT  LINE 

where  LK  cuts  the  axis  of  y.    If  OB  =  b,  the  coordinates  of  B 
are  (0,  b~).    Then  the  equation  of  LK  is 

y  —  b  —  7n  (x  —  0), 

or,  after  a  simple  reduction, 

y  =  mx  +  b.  (1) 

It  is%)  be  noticed  that  the  equation  of  a  line  parallel  to  the 
axis  of  y  cannot  be  put  in  this  form,  since  the  line  does  not  cut 
OY,  but  the  equation  of  any  other  line  can  be  given  this  form. 

Conversely,  any  equation  of  the  form  (1),  no  matter  what  are 
the  values  of  m  and  b,  represents  a  straight  line.  For  a  straight 
line  can  be  drawn  with  any  slope  m  and  any  intercept  b.  The 
equation  of  this  line  is  then  y  =  mx  +  b,  and  this  equation  is  sat- 
isfied by  no  point  not  on  the  line. 

30.  The  two-point  equation.  If  a  straight  line  is  determined 
by  the  two  points  L^(xv  y^)  and  B2(x2,  y2~),  then 

y-i  -  Vx 

m  = , 

x*~xi 

by  §  6,  and  the  equation  of  the  line  is  by  (1),  §  28, 

y-yx=i^rjix-xd'  CO 

If  y2  =  yv  the  line  is  parallel  to  OX,  and  its  equation  is 

y  =  y,  (2) 

If  x2  =  x^  the  line  is  parallel  to  0  Y,  and  its  equation  is 

x  =  x  .  (3) 

Ex.    Find  a  straight  line  through  (1,  2)  and  (—  3,  5). 
By  formula  (1),  p-  _  9 

or  3  x  +  4  y  -  11  =  0. 

31.  The  general  equation  of  the  first  degree.    The  equation 

Ax  +  By  +  C=0, 

where  A,  B,  and  C  may  be  any  numbers  or  zero,  except  that 
A  and  B  cannot  be  zero  at  the  same  time,  is  called  the  general 


ANGLES  59 


equation  of  the  first  degree.  We  shall  prove :  The  general 
equation  of  the  first  degree  with  real  coefficients  always  represents 
a  straight  line. 

1.  Suppose  A  =£  0  and  B  =£  0.    The  equation  may  be  written 

A         C 
&  B         B 

This  equation  is  of  the  form  y  =  mx  +  b  and  therefore  repre- 
sents a  straight  line,  by  §  29. 

It  follows  that  if  the  equation  of  a  straight  line  is  in  the  form 
Ax  +  By  +  C  =  0,  its  slope  may  be  found  by  solving  the  equation 
for  y  and  taking  the  coefficient  of  x. 

2.  Suppose  A  =  0,  B  =£  0.    The  equation  is  then 

(j 
By+C=0,     or     y  =  -~, 
B 

C 
and  represents  a  straight  line  parallel  to  OX  at  a  distance 

units  from  it. 

3.  Suppose  A  =£  0,  B  =  0.    The  equation  is  then 

Q 

Ax  +  C=0,     or     x  = , 

A 


and  represents  a  straight  line  parallel  to  O  Y  at  a  distance 

units  from  it. 

Therefore  the  equation  Ax  +  By  +  C=0  always  represents  a 
straight  line. 

32.  Angles.  The  slope  of  a  straight  line  enables  us  to  solve 
many  problems  relating  to  angles,  some  of  which  we  take  up  in 
this  article. 

1.  The  angle  between  the  axis  of  x 
and  a  known  line.  Let  a  known  line 
cut  the  axis  of  x  at  the  point  L. 
Then  there  are  four  angles  formed. 
To  avoid  ambiguity  we  shall  agree 
to  select  that  one  of  the  four  which 
is    above   the    axis    of   x   and   to  the  FlG"  66 

right  of  the  line  and  to  consider  LX  as  the  initial  line  of  this 
angle.    We  shall  denote  this  angle  by  $.    Then  if  we  take  any 


60 


THE  ST11A1G1IT  LINE 


point  P  on  the  terminal  line  of  <f>  and  drop  the  perpendicular 
J//',  we  have,  in  the  two  cases  represented  by  figs.  66  and  67, 


tan  (/> 


MP 
LM 


MP 


But  — —  is  equal  to  the  slope  of  the 
LM 

line,  by  (2),  §  6.    Therefore 
tan  d>  =  m. 


If  the  straight  line  is  parallel  to 
OY,  (/>  =  90°  and  tan  <f>  =  oo.  If  the 
line    is   parallel  to    OX,   no   angle   <f> 

is   formed,  but   since   m  =  0,   we   may   say   tan  <f>  =  0  ;    whence 
<£  =  0°  or  180°. 

2.  Parallel  lines.  If  two  lines  are  parallel,  they  make  equal 
angles  with  OX,  and  hence  their  slopes  are  equal.  It  follows  that 
two  equations  which  differ  only  in  the  absolute  term,  such  as 


Fig.  67 


and 


Ax  +  By  +  Cx  =  0 
Ax+By+C =0, 


represent  two  parallel  lines.    It  is  to  be  noticed  that  these  two 
equations  have  no  common  solution  (§14). 

Ex.  1.    Find  the  equation  of  a  straight  line  passing  through  (—  2,  3)  and 
parallel  to3x-5?/  +  6  =  0. 

First  method.    The  slope  of  the  given  line  is  %.    Therefore  the  required 

y  —  3  =  §  (a;  +  2),     or     3ar-5y  +  21  =  0. 

Sea/ml  method.   We  know  that  the  required  equation  is  of  the  form 

3x-5y  +  C  =  0, 

where  C  is  unknown.    Since  the  line  passes  through  (—  2,  3), 

3  (-2) -5  (3)  +  C=0, 

whence  C  =  21.    Therefore  the  required  equation  is 

3  x  —  5  y  +  21  =  0. 


ANGLES 


61 


3.  Perpendicular  lines.  Let  AB  and  CD  (fig.  68)  be  two  lines 
intersecting  at  right  angles.  Through  P  draw  PR  parallel  to 
OX,  and  let  RPD  =  </>1  and  RPB  =  <j>2.  Then  tan  </>1  =  my  and 
tan  <f>2  =  m2,  where  mx  and  ra2  are  the  slopes  of  the  lines.  But, 
by  hypothesis,  ^  =  ^+90°; 

whence  tan  <£2  =  —  cot  4>l 


which  is  the  same 


tan  <£x 


That  is,  two  straight  lines  are  perpendicular  when  the  slope  of 
one  is  minus  the  reciprocal  of  the  slope  of  the  other.  This  theorem 
may  be  otherwise  expressed 
by  saying  that  two  lines  are 
perpendicular  when  the  prod- 
uct of  their  slopes  is  minus 
unity. 

It  follows  that  two  straight 
lines  whose  equations  are  of 
the  type 

Ax+By  +  Cx  =  0 
and  Bx  -  Ay  +  C\=0 
are  perpendicular. 

Ex.  2.    Find  a  straight  line  through  (•">,  3)  perpendicular  to  7x+0y+l  =  0. 
First  method.   The  slope  of  the  given  line  is  —  l-   Therefore  the  slope 
of  the  required  line  is  £.    Therefore  the  required  line  is 

y  -  3  =  2  (a;  -  5),     or     9  x  -  7  y  -  24  =  0. 

Second  method.  We  know  that  the  equation  of  the  required  line  is  of 
the  form  Qx  —  7y  +  C  =  0.  Substituting  (5,  3),  we  find  C  =—  24.  Hence 
the  required  line  is  9  x  —  7  y  —  24  =  0. 

Ex.  3.  Find  the  equation  of  the  perpendicular  bisector  of  the  line  join- 
ing (0,  5)  and  (5,  — 11).  The  point  midway  between  the  given  points  is 
(§,  —  3),  by  §  7.  The  slope  of  the  line  joining  the  given  points  is  —  VS 
by  §  6.  Hence  the  required  line  passes  through  (§,  —  3),  with  the  slope  f^. 
Its  equation  is 

y  +  3  =  &  (x  -  |),      or      10  x  -  32  y  -  121  =  0. 


Fm 


G2 


THE  STRAIGHT  LINE 


4.  Angle  between  two  lines.  Let  AB  and  CD  (fig.  69)  inter- 
sect at  the  point  P,  making  the  angle  BPD,  which  we  shall 
call  /3.  Draw  the  line  PR  parallel  to  OX,  and  place  BPB  =  ^ 
andJSPjr>  =  £,    Then  0  =  ^-^; 

tan  (^ 


hence 


tan  /3  =  tan  (</>., 


_tan</> 
?V      1  + 


But  tan  <f>1 = ml  and  tan  c/>2 
=  m2,  where  ?n2  is  the  slope 
of  CD  and  w1  is  the  slope  of 
AB'.    Therefore 

tan  ft 


1  +  m^ 

If  <£>2  is  always  taken 
greater  than  c^,  tan  /3  will 
be  positive  or  negative 
according  as  /?  is  acute 
or  obtuse. 

Ex.  4.    Find  the  acute  angle  between  the  two  lines 

2a:-3?/  +  5  =  0     and     x  +  2y+2  =  0. 

Since  the    second   line    makes  the  larger   angle  with    OX,  we   jilace 
m%=-  |,  mx  =  $. 

Then,  by  substituting  in  the  formula, 

tan/3=~  *~J"=-  ? 


Here  /3  is  an  obtuse  angle,  and  the  supplementary  acute  angle  is  tan-1  J. 

Ex.  5.  Find  the  equation  of  a  straight  line  through  the  point  (—  2,  0), 
making  an  angle  tan-1  §  with  the  line  3x  +  4?/  +  G  =  0. 

Here  tan/3  is  given  as  §,  and  one  of  the  slopes  mz  or  mx  is  known  to 
be  —  ^.  Since  it  is  unknown  which  of  the  slopes  is  —  ^,  the  problem 
has  two  solutions : 

(1)  Place  m2  =—  |.    Then,  by  substituting  in  the  formula, 
2       -  |  -  to.         .  17 

-  =  - 1.     whence      m.  = . 

3 


1-flBj  *  ( 

The  equation  of  the  required  line  is  then 

y  -  0  =  -  V-  (x  +  2), 
17a:  +  6y  +  34  =  0. 


DISTANCE  FROM  A  STRAIGHT  LINE 


63 


(2)  Place 


Then 


whence 


The  equation  of  the  required  line  is  then 

or  x  +  18  y  +  2  =  0. 

33.  Distance  of  a  point  from  a  straight  line.    Let  LK  (fig.  70) 
be  a  given  straight  line  with  the  equation 

0, 


C: 


Ax+By 

and  let  Pl(xl,  y^)  be  a  given  point 
length  of  the  perpendicular  J^E 
drawn  from  1\  to  LK. 

Draw  the  ordinate  ML\  and 
let  it  intersect  the  line  LK  in 
the  point  Q.  Then  the  abscissa 
of  Q  is  a?,  and  its  ordinate  may 
be  denoted  by  y2.  Since  Q  is  on 
the  line  LK.  we  have 


It  is  required  to  find  the 

F 


Axx+By%+C=0, 


Fig.  70 


whence 


Then 


OT-Jfi-Jf, 


J./-,  +  />'//,  +  (? 


It  is  clear  that  this  expression  is  a  positive  quantity  when 
(xlt  y^)  lies  above  the  line  LK  and  is  a  negative  quantity  when 
(x^  y^)  lies  below  ZA".  It  is  also  evident  from  the  triangle  T[QB, 
and  from  a  like  triangle  in  other  cases,  that  the  length  of  P^  is 

A 
numerically  equal  to  QPX  cos  cj>.    But  tan  £  =  —  —  i  and  hence 

B 

B 


We  have,  then, 


cos  <j)  = 


Pi? 


±V.I2  +  Z2 
J^+Ti^  +  C 


64  THE  STRAIGHT  LINE 

We  may,  if  we  wish,  always  choose  the  4-  sign  in  the  denomi- 
nator. Then  PR  is  positive  for  all  points  on  one  side  of  the  line 
Ax  -f-  By  +  C  =  0  and  negative  for  all  points  on  the  other  side. 
To  determine  which  side  of  the  line  corresponds  to  the  positive 
sign,  it  is  most  convenient  to  test  some  one  point,  preferably 
the  origin. 

Ex.  Find  the  distance  from  the  point  (7,  —  4)  to  the  line  2  x  +  3  y 
+  8  =  0. 

By  use  of  the  formula, 

p^_2(7)+3(-4)+8  =    10_ 

Via  V13 

Since  the  coordinates  of  the  origin,  similarly  substituted,  give  a  positive 
sign  to  the  result,  the  point  (7,  —  4)  is  on  the  same  side  of  the  line  as  the 
origin.    A  plot  verifies  this. 

PROBLEMS 

1.  Find  the  equation  of  the  straight  line  passing  through  (1,  —  3) 
with  the  slope  2. 

2.  Find  the  equation  of  the  straight  line  passing  through  (— 1, 
—  ^)  with  the  slope  —  3. 

3.  Find  the  equation  of  the  straight  line  passing  through  (5,  —1) 
with  its  slope  the  same  as  that  of  the  straight  line  determined  by 
(0,  3)  and  (2,  0). 

4.  Find  the  equation  of  the  straight  line  passing  through  (2,  —  f) 
with  the  slope  zero. 

5.  Find  the  equation  of  the  straight  line  passing  through  (^,  §) 
with  an  infinite  slope. 

6.  Find  the  equation  of  the  straight  line  of  which  the  slope  is  5 
and  the  intercept  on  OY  is  —  4. 

7 .  Find  the  equation  of  the  straight  line  of  which  the  slope  is  —  3 
and  the  intercept  on  OF  is  \. 

8.  Find  the  equation  of  the  straight  line  of  which  the  slope  is  0 
and  the  intercept  on  OY  is  —  §. 

9.  Find  the  equation  of  the  straight  line  through  the  points 
(_1,  _4)  and  (0,  5). 

10.  Find  the  equation  of  the  straight  line  through  the  points 
(2,-1)  and  {-1,1). 


PROBLEMS  65 

11.  Find  the  equation  of  the  straight  line  through  the  points 
(2,  -1)  and  (2,  3). 

12.  What  is  the  equation  of  a  straight  line  the  intercepts  of  which 
on  the  axes  of  x  and  y  are  3  and  —  4  respectively  ? 

13.  What  is  the  equation  of  the  straight  line  the  intercepts  of 
which  on  the  axes  of  x  and  y  are  —  5  and  —  8  respectively  ? 

14.  Derive  the  equation  of  the  straight  line  the  intercepts  of 
which  on  the  axes  of  x  and  y  are  a  and  b  respectively. 

15.  Find  the  equation  of  a  straight  line  through  (|,  §)  and  the 
point  of  intersection  of  the  lines  3x  —  5y— 11  =  0  and  4 x  +  y  —  7  =  0. 

16.  Find  the  equation  of  the  straight  line  joining  the  point  of  inter- 
section of  the  lines  2x  —  y  — 1=0  and  x  —  y  4-  7  =  0  and  the  point 
of  intersection  of  the  lines  x  —  1  y  — 1=0  and  2x  —  5  y  +1=  0. 

17.  Find  the  equation  of  the  straight  line  passing  through  (2,  —  3) 
and  making  an  angle  of  120°  with  OX. 

18.  Find  the  equation  of  the  straight  line  making  an  angle  of  30° 
with  OX  and  cutting  oft'  an  intercept  3  on  0  >'. 

19.  A  straight  line  making  a  zero  intercept  on  OF  makes  an  angle 
of  45°  with  OX.    Find  its  equation.  Y^ 

20.  A  straight  line  making  a  zero  angle  with  OX  cuts  OF  at  a 
point  3  units  from  the  origin.    Find  its  equation. 

21.  Find  the  equation  of  the  straight  line  through  (2,  —  3)  parallel 
to  the  line  2  x  4-  y  =  7. 

22.  Find  the  equation  of  the  straight  line  through  (—  |,  —  2) 
parallel  to  the  line  3x  —  2^4-2  =  0. 

23.  Find  the  equation  of  the  straight  line  passing  through  (—  1,  —  1) 
parallel  to  the  straight  line  determined  by  (—  2,  G)  and  (2,  1). 

24.  In  the  triangle  A  (—2,  -1),  5(3,  1),  C(—  1,  4)  a  straight 
line  is  drawn  bisecting  the  adjacent  sides  AB  and  BC.  Prove  by 
computation  that  it  is  parallel  to  AC  and  half  as  long. 

25.  Find  the  equation  of  the  straight  line  passing  through  the 
point  of  intersection  of  x  —  3y  +  2  =  0  and  5x  -\-  6y  —  4  =  0  and 
parallel  to  4x  +  ?/4-7=0. 

26.  Find  the  equation  of  the  straight  line  parallel  to  the  line 
a-f-3y  —  5  =  0  and  bisecting  the  straight  line  joining  (—  2,  —  3) 
and  (5,  5). 


M  THE  STRAIGHT  LINE 

27.  Find  the  equation  of  the  straight  line  through  the  origin 
perpendicular  to  the  line  3  a-  +  4  «/  — 1=0. 

28.  Find  the  equation  of  the  straight  line  through  (2,  —  3)  per- 
pendicular to  the  line  7  x  —  4  y  +  3  =  0. 

29.  Find  the  equation  of  the  perpendicular  bisector  of  the  straight 
line  joining  the  points  (—  5,  —1)  and  (—  3,  4). 

30.  A  straight  line  is  perpendicular  to  the  line  joining  the  points 
(_  4}  0)  and  (4,  —  1)  at  a  point  one  third  of  the  distance  from  the 
first  point  to  the  second.    What  is  its  equation  ? 

31.  Find  the  equation  of  the  straight  line  perpendicular  to 
2x  —  3  y  -f-  7  =  0  and  bisecting  that  portion  of  it  which  is  included 
between  the  coordinate  axes. 

32.  Find  the  equation  of  the  straight  line  through  the  point  of 
intersection  of  6x  —  2  ?/  +  8  =  0  and  4a;-  6y  -\-  3  =  0  and  per- 
pendicular to  5x  +  2y  +  6  =  0. 

33.  Find  the  equation  of  the  perpendicular  bisector  of  the  base  of 
an  isosceles  triangle  having  its  vertices  at  the  points  (4,  3),  (—  1,  —  2), 
and  (3,  -  4). 

34.  Find  the  acute  angle  between  the  lines  x  —  y  -\-  4  =  0  and 
3x-y  +  6  =  0. 

35.  Find  the  acute  angle  between  the  lines  2x  —  y  +  8  =  0  and 
2x  +  5y-4  =  0. 

36.  Find  the  acute  angle  between  the  lines  x  +  y  —  5  =  0  and 
4x  +  y-8  =  0. 

37.  Find  the  acute  angle  between  the  line  3x  —  2 y  +  6  =  0  and 
the  line  joining  (4,  —  5)  and  (—  3,  2). 

38.  Find  the  acute  angle  between  the  straight  lines  drawn 
from  the  origin  to  the  points  of  trisection  of  that  part  of  the  line 
2x  -\-  3y  — 12  =  0  which  is  included  between  the  coordinate  axes. 

39.  Show  that  x  —  y  +  3  =  0  bisects  one  of  the  angles  between 
the  lines  4x-3v/+ll=0  and  3 x  —  4 y '+ 10  =  0. 

40.  Find  the  vertices  and  the  angles  of  the  triangle  formed  by 
the  lines  3a:  +  5?/^14  =  0,  9 a;  —  ?/  +  22  =  0,  and  x  —  y  —  2  =  0 

41.  Find  the  equations  of  the  straight  lines  through  the  point 
(—  3,  0)  making  an  angle  tan_1i  with  the  line  3x  —  5y  +  9  =  0. 

42.  Find  the  equations  of  the  straight  lines  through  (4,  —3) 
making  an  angle  of  45°  with  the  line  3  x  +  4  y  =  0. 


PKOBLEMS  67 

43.  Find  the  equations  of  the  straight  lines  through  the  point 
(—  1,  —  1)  making  an  angle  tan-1  \  with  the  line  3x-\-2y  —  6  =  0. 

44.  Find  the  equations  of  the  straight  lines  through  the  point 
(2, 1)  making  an  angle  tan-1 2  with  the  line  2x  —  y  +  4  =  0. 

45.  Find  the  equations  of  the  straight  lines  through  the  point 
(3, 1)  making  an  angle  tan-1 3  with  the  line  x  +  3y  —  3  =  0. 

46.  Find  the  distance  of  (2, 1)  from  the  line  y  =  3x  -\-7. 

47.  Find  the  distance  of  (2,  —  f)  from  the  line  x  -{-  2  y  —  4  =  0. 

48.  Find  the  distance  of  the  point  (b,  —  a)  from  the  line 
bx  -f-  ay  =  ah. 

49.  The  equations  of  the  sides  of  a  triangle  are  respectively 
3 x  +  5 y ■  —  16  =  0,  x  —  y  =  0,  and  3cc  +  y  +  4  =  0.  Find  the  dis- 
tance of  each  vertex  from  the  opposite  side. 

50.  The  base  of  a  triangle  is  the  straight  line  joining  the  points 
(—  3, 1)  and  (5,  —  1).  How  far  is  the  third  vertex  (6, 5)  from  the  base  ? 

51.  The  vertex  of  a  triangle  is  the  point  (5,  3),  and  the  base  is 
the  straight  line  joining  (—  2,  2)  and  (3,  —  4).  Find  the  lengths  of 
the  base  and  the  altitude. 

52.  Find  the  equations  of  the  medians  of  the  triangle  formed  by 
the  lines  2x  -  3y  4-  11  =  0,  3*  +  y  -  11  =  0,  and  x  +  Ay  =  0. 

53.  Find  the  foot  of  the  perpendicular  drawn  from  the  point 
(-  1,  2)  to  the  line  3x-5y-21  =  0. 

54.  Find  the  distance  between  the  two  parallel  lines  2x+3y— 8=0 
and  2x  +  3y-  10  =  0. 

55.  Find  the  distance  between  the  two  parallel  lines  3x—5y+l  =  Q 
and3x-5y-7  =  0. 

56.  A  triangle  has  the  vertices  (2,  4),  (3,  —1),  and  (—  5,  3).  Find 
the  distance  from  the  vertex  (2,  4)  to  the  point  of  intersection  of 
the  median  lines. 

57.  A  straight  line  is  drawn  through  (2,  —  3)  perpendicular  to  the 
line  3x  —  4  ?/  4-  6  =  0.    How  near  does  it  pass  to  the  point  (6,  8)  ? 

58.  Determine  the  value  of  m  so  that  the  line  y  =  mx  4-  3  shall 
pass  through  the  point  of  intersection  of  the  lines  y  =  2x  +  l  and 
y  =  x  4-  5. 

59.  A  straight  line  passes  through  the  point  (—  \,  4),  and  its 
nearest  distance  to  the  origin  is  2  units.    What  is  its  slope  ? 


08  THE  STKAIGHT  LINE 

60.  One  diagonal  of  a  parallelogram  joins  the  points  (3,  —  1)  and 
(—3,  —  3).  One  end  of  the  other  diagonal  is  (2,  3).  Find  its  equation 
and  its  length. 

61.  Perpendiculars  are  let  fall  from  the  point  (9,  5)  upon  the 
sides  of  the  triangle  the  vertices  of  which  are  at  the  points  (8,  8), 
(0,  8),  and  (4,  0).  Show  that  the  feet  of  the  three  perpendiculars 
lie  on  a  straight  line. 

62.  Find  a  point  on  the  line  2x  -\-3y  —  6  =  0  equidistant  from 
the  points  (4,  4)  and  (6,  1). 

63.  Find  a  point  on  the  line  5x  —  3?/  +  15  =  0  the  distance  of 
which  from  the  axis  of  x  equals  §  its  distance  from  the  axis  of  y. 

64.  A  point  is  equally  distant  from  (3,  2)  and  (—  3,  4),  and  the 
slope  of  the  straight  line  joining  it  to  the  origin  is  $.  Where  is  the 
point  ? 

65.  A  point  is  8  units  distant  from  the  origin,  and  the  slope  of  the 
straight  line  joining  it  to  the  origin  is  —  \.   What  are  its  coordinates  ? 

66.  A  point  is  5  units  distant  from  the  point  (1,  —  2),  and  the 
slope  of  the  line  joining  it  to  (0,  —  8)  is  \.    Find  the  point. 

67.  Find  the  points  on  the  straight  line  determined  by  (1, 1)  and 
(—  2,  —  3)  which  are  15  units  distant  from  either  of  the  given  points. 

68.  Prove  analytically  that  the  locus  of  points  equally  distant 
from  two  points  is  the  perpendicular  bisector  of  the  straight  line 
joining  them. 

69.  Prove  analytically  that  the  medians  of  a  triangle  meet  in  a 
point. 

70.  Prove  analytically  that  the  perpendiculars  from  the  vertices 
of  a  triangle  to  the  opposite  sides  meet  in  a  point. 

71.  Prove  analytically  that  the  straight  lines  joining  the  middle 
points  of  the  adjacent  sides  of  any  quadrilateral  form  a  parallelogram. 

72.  Prove  analytically  that  the  perpendicular  bisectors  of  the  sides 
of  a  triangle  meet  in  a  point. 

73.  Prove  analytically  that  the  perpendiculars  from  any  two  ver- 
tices of  a  triangle  to  the  median  from  the  third  vertex  are  equal. 

74.  Prove  analytically  that  the  straight  lines  drawn  from  a  vertex 
of  a  parallelogram  to  the  middle  points  of  the  opposite  sides  trisect 
a  diagonal. 


CHAPTER  VI 

CERTAIN  CURVES 

34.  Locus  problems.  A  curve  is  often  defined  as  the  locus  of 
a  point  which  has  a  certain  geometric  property.  It  is  then  usually 
possible  to  obtain  the  equation  of  the  curve  by  expressing  this 
property  by  means  of  an  equation  involving  the  coordinates  of 
any  point  of  the  locus.  This  is  illustrated  in  the  following 
examples : 

Ex.  1.  Find  the  locus  of  a  point  at  a  distance  3  from  the  straight  line 
ix+By  —  6  =  0. 

Let  (x,  y)  be  any  point  of  the  locus.    By  §  33,  the  distance  of  (a?,  ?/)  from 

the  given  straight  line  is  ± '— Hence,  by  the  conditiun.s  of  the 

problem, 

4  *  +  3  //  -  0  _ 

•"> 

which  reduces  to    4x  +  3y  — 21  =  0,    or     1  .,+:'.//  + 0  =  0. 

These  are  the  equations  of  two  straight  lines  parallel  to  the  given  line. 

Ex.  2.  Find  the  locus  of  a  point  at  ;i  distance  9  from  the  point 
(-  5,  -  3). 

Let  (x,  j/)  be  any  point  of  the  locus.  Its  distance  from  (—5,  —  •"> )  is, 
by  §  5,  V(x  +  5)a  +  (//  +  '■'>)';.    Hence,  by  the  conditions  of  the  problem, 


V(x+  ."')-  +  (//  +  :'.)- =  n, 
which  reduces  to        x2  +  if  +  K)  x  +  (i  //  —  47  =  0. 

This  is  the  equation  of  the  required  locus.  The  curve  may  be  plotted 
from  the  equation  or  may  be  drawn  with  compasses,  as  it  is  obviously 

a  circle. 

In  the  following  articles  we  shall  employ  the  methods  just 
illustrated,  to  obtain  the  equations  of  certain  important  curves. 
An  equation  thus  obtained  may  be  used  both  for  plotting  the 
curve  and  for  examining  its  properties. 


70 


CEKTAIN  CURVES 


35.  The  circle.  A  circle  is  the  locus  of  a  point  at  a  constant 
distance  from  a  fixed  point.  The  fixed  point  is  the  center  of  the 
circle,  and  the  constant  distance  is  the  radius. 

Let  (7i,  &)  be  the  center  C  (fig.  71),  and  let  r  be  the  radius  of 
the  circle.  Then  if  P  (x,  y~)  is  a  point  on  the  circle,  x  and  y  must 
satisfy  the  equation  • 

(x-hy+(y-ky=r\    (1) 

by  §5. 

Conversely,  if  x  and  y  satisfy 
the  equation  (1),  the  point 
(x,  y~)  is  at  a  distance  r  from 
(A,  k)  and  therefore  lies  on  the 
circle. 

Therefore  (1)  is  the  equation 
of  the  circle. 

Equation  (1)  expanded  gives 

x2  +y2-2  hx  -  2  Icy  +  Jr  +  P 
and  if  this  is  multiplied  by  any  quantity  A,  it  becomes 
Ax2  +  Ay'2  +  2  Gx  +2Fy  +  C=0, 


(2) 


where 


tf+k2-r2  = 


Ex.    The  equation  of  a  circle  with  the  center  (J,  —  ^)  and  the  radius  §  is 

which  reduces  to        12  x2  +  12  y"  —  12  x  +  8  y  —  1  =  0. 

36.   Conversely,  the  equation 

Ax2+Ay2-\-  2  Gx  +  2Fy  +  C=  0, 

where  A  =#=  0,  represents  a  circle  if  it  represents  any  curve  at  all. 
To  prove  this  we  will  follow  the  method  of  Ex.  2,  §  18,  and 
write  the  equation  in  the  form 


<"+SM'+3'- 


F2-AC 


THE  CIRCLE  71 

There  are  then  three  possible  cases : 

1.  G2  +  F2-AC>0.    The  equation  is  then  of  the  type  (1), 
§  35,  where  h  = ,  k  = ,  r  =  — — - ,  and  therefore 

represents  a  circle  with  the  center  ( -, )  and  the  radius 

r-? 5 \     A        A 

G2+F2-AC 

\ A 


2.   G2+F2-AC=0.    The  equation  is  then 


which  can  be  satisfied  by  real  values  of  x  and  y  only  when 

G                     F 
x  — and  y  = Hence  the  equation  represents  the  point 

('        F\ 

— , J.   This  may  be  called  a  circle  of  zero  radius,  regarding 

it  as  the  limit  of  a  circle  as  the  radius  approaches  zero. 

3.  G2  +  F2  —  A C  <  0.  The  equation  can  then  be  satisfied 
by  no  real  values  of  x  and  y,  since  the  sum  of  two  positive 
quantities  cannot  be  negative.  Hence  the  equation  represents 
no  curve. 

Ex.  1.    The  equation  2  x1  +  -2  >/-  +  •-'  x  -  '2  y  -  .">  =  0  may  be  written 

(,■+   \f  +  {!l-  lf  =  Z, 

and  represents  a  circle  with  the  center  (—  .1.  \ )  ;tn«l  the  radius  ^  :!.  Tins 
circle  can  now  be  drawn  with  compasses,  the  methods  of  Chapter  II  not 
being  required. 

Ex.  2.    The  equation  x*  +  //-  —  2  x  +  1  y  +  5  =  0  may  be  written 
(ar-l)s+(y  +  2)2  =  0, 

and  is  satisfied  only  by  the  point  (1,  —  2). 

Ex.  3.    The  equation  .c2  +  y*  —  2  .*•  +  4  //  +  7  =  0  may  be  written 
(,:-l)2  +  (//  +  2)2=-2, 
and  represents  no  curve. 


72  CERTAIN  CURVES 

37.  To  find  '  equation  of  a  circle  which  will  satisfy  given 
conditions,  it  necessary  and  sufficient  to  determine  the  three 
quantities  h,  k,  r,  or  the  ratios  of  the  four  quantities  A,  G,  F,  C. 
Each  condition  imposed  upon  the  circle  leads  usually  to  an  equa- 
tion involving,  these  quantities.  In  order  to  determine  the  three 
quantities  J.-  ..  necessary  and  in  general  sufficient  to  have  three 
equations.  Henc  in  general,  three  conditions  are  necessary  and 
sufficient  U.  ;iine  a  circle. 

It  is  l  tant  to  enumerate  all  possible  conditions  which 

may  be  ..^  id  upon  a  circle,  but  the  following  three  may  be 
menti< 

1.  ?£ftt  the  condition  be  imposed  upon  the  circle  to  pass 
through  the  known  point  (x,  y^).  Then  (xx,  y^)  must  satisfy 
the  equation  of  the  circle ;  therefore  h,  k,  and  r  must  satisfy  the 
condition 

QXl~hy+(y-ky  =  r\ 

2.  Let  the  condition  be  imposed  upon  the  circle  to  be  tangent 
to  the  known  straight  line  Ax  +  By  +  C  =  0.  Then  the  distance 
fro.n  the  center  of  the  circle  to  this  line  must  equal  the  radius; 
therefore,  by  §  33,  h,  k,  and  r  must  satisfy  the  condition. 

Ah+Bk  +  C 

=±r. 

y/A2+B2 

The  sign  will  be  ambiguous  unless  from  other  conditions  of  the 
problem  it  is  known  on  which  side  of  the  line  the  center  lies. 

3.  Let  it  be  required  that  the  center  of  the  circle  should  lie  on 
the  line  Ax  +  By  +  G  =  0.   Then  h  and  k  must  satisfy  the  condition 

A7i  +  Bk+G=0. 

Ex.  1.  Find  the  equation  of  the  circle  through  the  three  points  (2,  —  2), 
(7,  3),  and  (G,  0). 

The  quantities  h,  k,  and  r  must  satisfy  the  three  conditions 


0 

l-i 

'02  + 

(-2- 

-  ^')2 

=  r2, 

(7- 

-]<y 

+  (S- 

-  A-)2 

=  r\ 

(6- 

-  >02 

+  ro- 

-  l:f 

=  r\ 

THE  CIRCLE  73 

Solving  these,  we  have  ft  =  2,  k  =  3,  and  r  =  5.       '      efore  the  required 
equation  is 

(x  -  2)2  +  (y  -  3)2  =  25, 

or  x2  +  #2  —  4  x  —  6  y  —  12  =  0. 

Ex.  2.    Find  the  equation  of  the  circle  which  passes  4l  — >ugh  the  points 
(2,  —  3)  and  (—  4,  —  1)  and  has  its  center  on  the  line  3  j  n  x  —  18  =  0. 
The  quantities  ft,  k,  and  r  must  satisfy  the  condi  i'>ns 

(2  -  7,)2  +  (-  3  -  kf  =  r2, 

(_.l_/,)2  +  (_l_  J.)2  =  ,.2; 

3  A-  +  h  -  18  =  0. 

Solving  these  equations,  we  find  ft  =  |,  /.■  =  'J,  /•'- =  J -| "'.       herefore 
the  required  equation  is 

(a?_§)8  +  (y_J3L)a  =  xjAj 

or  z2  +  //2  -  3  x  — 11  y  -  40  =  0. 

Ex.  3.   Find  the  equation  of  a  circle  which  is  tangent  to  the  lines 

17  x  +  y  -  85  =  0     and     13  a;  + 11  y  +  50  =  0, 

and  has  its  center  on  the  line  88  x  +  70  y  +  15  =  0. 
The  quantities  ft,  k,  and  r  must  satisfy  the  conditions 

17*  +  *  -35 

=  ±  r, 


V290 
13  ft  +  11  jfc  +  50 


±r, 


a  290 

88  ft  +  70  A-  +  15  =  0. 

These  equations  have  the  two  solutions 

,_.,  ,        ,  =V200. 

,.  .       _  .  ,.,  3V290 

and  «  =  0,  K  =  — 'tj  ,         r  =  — • 

20 

Hence  each  of  the  two  circles 

3  x-  +  3  //-  +  5  x  -  5  y  -  20  =  0 

and  40  x2  +  40  if  -  400  x  +  520  y  +  2429  =  0 

satisfies  the  conditions  of  the  problem. 


74 


CERTAIN  CURVES 


38.   The  ellipse.    An  ellipse  is  the  locus  of  a  point  the  sum  of 
the  distances  of  which  from  two  fixed  points  is  constant. 

The  two  fixed  points  are  called  the  fori.  Let  them  be  denoted 
by  F  and  F'  (fig.  72),  and  let  the  axis  of  x  be  taken  through 
them,  and  the  origin  halfway 
between  them.  Then  if  F  is 
any  point  on  the  ellipse  and 
2  a  represents  the  constant 
sum  of  its  distances  from  the 
foci,  we  have 

F'P+FP=2a.       (1) 

From  the  triangle  F'PF  it 
follows  that 

F'F<2a. 

Hence  there  is  a  point  A  on  the  axis  of  x  and  to  the  right 
of  F  which  satisfies  the  definition.    We  have,  then, 

F'A+FA  =  2a, 
or  (F' 0  +  OA)  +  (OA-  OF)  =2a; 


whence 

Let  us  now  place 


0A  =  a. 

OF 

OA  =  e> 


diere  e<l. 


The  quantity  e  is  called  the  eccentricity  of  the  ellipse. 
Then  the  points  F  and  F'  are  (±  ae,  0).   Computing  the  values 
of  F'P  and  FP  by  §  5  and  substituting  in  (1),  we  have 


y/(x  +  aef+  y*+v(x  —  ae~)2+  y* 


(2) 


By  transposing  the  second  radical  to  the  right-hand  side  of  the 
equation,  squaring,  and  reducing,  we  have 


a  -  ex  =V(x  -  ae)2  +f  =  FP.  (3) 

Similarly,  by  transposing  the  first  radical  in  (2),  we  have 

a  +  ex  =  V(a;  +  ag)2+  f  =  F'P.  (4) 


THE  ELLIPSE  75 

Either  (3)  or  (4)  leads  to  the  equation 

(l-e>)r+f  =  cc(l-e%  (5) 

or  ~+        *    .    =1.  (6) 

a-      a"(l—  e-)  v  y 

Since  e  <  1,  the  denominator  of  the  second  fraction  is  positive, 
and  we  place 

a;2      ?/2 
thus  obtaining  —  +  %-  =  1.  (8) 

We  have  now  shown  that  any  pqint  which  satisfies  ( 1  )  has 

coordinates  which  satisfy  (8). 

We  may  show,  conversely,  that  any  point  whose  coordinates  satisfy  (8) 
is  such  as  to  satisfy  (1).  Let  us  assume  (8)  as  given.  We  can  then  obtain 
(G)  and  (">),  and  (5)  may  be  put  in  each  of  the  two  forms 

x2  +  2  aex  +  a2e2  +  y2  =  a2  +  2  aex  +  t  '-'.<-, 
x2  —  2  aex  +  ore2  +  y2  =  a9  —  -  cu  x  +  t '-'. i  -, 
the  square  roots  of  which  are  respectively 

F'P  =  ±  («  +  ex), 

FP  =  ±  (o  -  ex). 

These  lead  to  one  of  the  four  following  equations: 

/■"/■  +  FP  =  2a, 

F'P-  FP=  2a, 

-F'P+  FP=  2  a, 

-F'P-  FP=  2a. 

Of  these,  the  last  one  is  impossible,  since  the  sum  of  two  negative  num- 
bers cannot  be  positive;  and  the  second  and  third  are  impossible,  since  the 
difference  between  FP  and  F'P  must  be  less  than  F'F,  which  is  less  than  2  a. 
Hence  any  point  which  satisfies  (8)  satisfies  (1),  and  therefore  (8)  is  the 
equation  of  the  ellipse. 

39.  Placing  y  =  0  in  (8),  §  38,  we  find  x  =  ±a.  Placing  x  =  0, 
we  find  y  =  ±b.  Hence  the  ellipse  intersects  OX  in  two  points, 
A  (a,  0)  and  A'(—  a,  0),  and  intersects  OF  in  two  points,  B(Q,  b~) 
and  B'(0,  —  6).   The  points  A  and  A'  are  called  the  vertices  of  the 


76 


CERTAIN   CUEVES 


ellipse.  The  line  AA',  which  is  equal  to  2  a,  is  called  the  major 
axis  of  the  ellipse,  and  the  line  BB',  which  is  equal  to  2  b,  is 
called  the  minor  axis. 

Solving  (8)  first  for  y  and  then  for  x,  we  have 


=  +  -  Var- 


an d 


a 

a    /m 2 


These   equations   show  that  the  ellipse  is  symmetrical  with 
respect  to  both  OX  and  OY,  that  x  can  have  no  value  numeri- 
cally greater  than  a,  and  that  y  can  have  no  value  numerically 
greater  than  b.  If  we  construct 
the  rectangle  KLMN  (fig.  73), 
which  has  0  for  a  center  and 
sides    equal    to    2  a    and    2  b 
respectively,   the    ellipse   will 
lie  entirely  within  it;   and  if 
the  curve  is  constructed  in  one 
quadrant,  it  can  be  found  by 
symmetry    in    all    quadrants. 
The  form  of  the  curve  is  shown  Fig.  73 

in  figs.  72  and  73. 

40.  Any  equation  of  the  form  (8),  §  38,  in  which  a  >  b,  repre- 
sents an  ellipse  with  the  foci  on  OX.  For  if  we  place,  as  in  §  38, 
b2  =  a2(l—  e2),  we  find,  for  the  eccentricity  of  the  ellipse, 

Va^b2 


M 

B                        L 

A' 

v^ 

0                        J 

A    X 

1 

V 

B'                      K 

and  may  fix  F  and  F\  which  in  §  38  were  arbitrary  in  position, 
by  the  relation  OF  =  —  OF'  =  ae. 

The  foci  may  be  found  graphically  by  placing  the  point  of  a  com- 
pass on  B  and  describing  an  arc  with  the  radius  a.  This  arc  will 
intersect  A  A'  in  the  foci ;  for  since  OB=b  and  OF=  Va'2—  b2,  BF=  a. 

It  may  be  noted  that  the  nearer  the  foci  are  taken  together,  the 
smaller  is  e  and  the  more  nearly  b  =  a.  Hence  a  circle  may  be 
considered  as  an  ellipse  with  coincident  foci  and  equal  axes. 


THE  HYPERBOLA  77 

Similarly,  an  equation  of  the  form  (8),  §  38,  in  which  b  >  a, 
represents  an  ellipse  in  which  the  foci  he  on  OY  at  a  distance 
Vb2  —  a'2  from  0.  In  this  case  BB'  =  2  b  is  the  major  axis  and 
A  A'  =  2  a  is  the  minor  axis. 

Finally,  any  equation  of  the  form 

a2 

represents  an  ellipse  with  its  center  at  the  point  (li,  Tc)  and  its 
axes  parallel  to  OX  and  OY  respectively;  for  if  the  axes  arc 
shifted  to  a  new  origin  at  (//,  /r)  by  the  formulas  of  £17,  this 
equation  assumes  the  form  (8),  §  38. 

Ex.  1.  Show  that  4  x2  +  0  if  +  4  x  -  12  y  -  1  =  0  is  the  equation  of  an 
ellipse,  and  find  its  center,  semiaxes,  and  eccentricity. 

Following  the  method  of  Ex.  2,  §  18,  we  may  write  the  equation  in 

thef°rm  Hx+W  +  6(y-iy  =  8, 

3 

Hence  this  curve  is  an  ellipse  with  its  center  at  (—  \,  1)  and  its  major 

/-         4  Vij  1 

and  minor  axes  equal  respectively  to  2  V 2  and  — - —    Its  eccentricity  is  — —• 

'■''  vy 

Ex.  2.  Find  the  equation  of  an  ellipse  with  the  eccentricity  J  and  its 
foci  at  the  points  (—  1,  4),  (7,  1). 

Since  the  center  is  halfway  between  the  foci,  tin-  center  is  the  point 
(3,  4).  The  major  axis  of  the  ellipse  is  parallel  t<>  OX,  Bince  it  contains 
the  foci.    Since  each  focus  is  at  a  distance  ae  from  the  center, 

ae  =  4. 
But  e  =  J,  therefore  a  =  12. 

Then,  from  (7),  §  38,  b-  =  a-  (1  -  e2)  =  128. 

The  equation  of  the  ellipse  is  therefore 

Qr-3)2       Q/-4)2 
144  128 

which  reduces  to     8  x2  +  9  y"  —  48  x  —  72  y  —  936  =  0. 

41.  The  hyperbola.  An  hyperbola  is  the  locus  of  a  point  the 
difference  of  the  distances  of  which  from  two  fixed  ptoints  is  constant. 


78 


CERTAIN  CUEVES 


The  two  fixed  points  are  called  the  foci.  Let  them  be  F  and 
F'  (fig.  74),  and  let  FF'  be  taken  as  the  axis  of  x,  the  origin 
being  halfway  between  F  and  F'.  Then  if  P  is  any  point  on 
the  hyperbola  and  2  a  is  the  constant  difference  of  its  distances 
from  F  and  F'>  we  have 
either 


F'P 
FP- 


-FP=2a     (1) 
F'P  =  2  a.    (2) 


Fig.  74 


Since  in  the  triangle 
F'PF  the  difference  of  the 
two  sides  FP  and  F'P  is 
less  than  F'F,  it  follows 
that  F'F  >2a. 

There  is  therefore  at  least  one  point  A  between   0  and  F 
which  satisfies  the  definition. 

Then  F'A-AF=2a, 

or  (F'0  +  OA)-(OF-OA-)=2a; 

whence  OA  =  a. 

We  may  therefore  place 

. —  =  e.  where  e>l. 

OA 

The  quantity  e  is  called  the  eccentricity  of  the  hyperbola. 
Then  the  points  F  and  F'  are  (±  ae,  0),  and  equations  (1) 
and  (2)  become 


and 


^/(x  +  aey+if  -  y/(x  -  aef  +tf=2a 
^/(x-aey+y1-  yf(x  +  ae)'2+  f  =2  a. 


(3) 
(4) 


By  transposing  one  of  the  radicals  to  the  right-hand  side  of 
these  •  equations,  squaring,  and  reducing,  we  obtain  from  either 


(3)  or  (4) 


(l-_es)s3+/  =  a*(l_e3), 


dz  +  d2(l-e~) 


(5) 
(<0 


THE  HYPERBOLA 


79 


But  since  e>l,   <r(l  —  e~)   is  a  negative  quantity,   and  w 


may  write  a2  (1  —  e2)  =  —  b'2,  thus  obtaining 


C=i, 


(?) 


an  equation    satisfied   by  the  coordinates  of   any  point   which 
satisfies  (1). 

Proceeding  as  in  §  38,  we  may  prove,  conversely,  that  any  point  whose 
coordinates  satisfy  (7)  is  such  as  to  satisfy  either  (1)  or  (2),  and  hence  is 
a  point  of  the  hyperbola. 

42.  If  we  place  y  =  0  in  (7),  §41,  we  have  x  =  ±a.  Hence 
the  curve  intersects  OXva.  two  points,  A  and  ./',  called  the  vertices. 
If  x  =  0,  y  is  imaginary.    Hence  the  curve  does  not  intersect  0  Y. 

Solving  (7),  §  41,  for  y  and  x  respectively,  we  have 


y  =  ±-  va 


and 


These  show  that  the  curve  is 
symmetrical  with  respect  to  both 
OX  and  OY,  that  x  can  have  no 
value  numerically  less  than  a,  and 
that  y  can  have  all  values. 

Moreover,  the  equation  for  y  can 
be  written 


a     N        :r 


Fig.  75 


As  rr  increases,  the  term  —  decreases,  approaching  zero  as  a 

x' 

limit.    Hence  the  more  the  hyperbola  is  prolonged,  the  nearer  it 

comes  to  the  straight  lines  y  =  ±-x.    Therefore  the  straight 

lines  y  =  ±-x  are  the  asymptotes  of  the  hyperbola.    They  are 

the  diagonals  of  the  rectangle  constructed  as  in  fig.  75  and  are 
used  conveniently  as  guides  in  drawing  the  curve.    The  line  AA' 


80  CERTAIN  CURVES 

is  called  the  transverse  axis,  and  the  line  BB'  the  conjugate  axis, 
of  the  hyperbola.  The  shape  of  the  curve  is  shown  in  figs.  74 
and  75. 

43.  Any  equation  of  the  form  (7),  §  41,  where  a  and  b  are 
any  positive  real  values,  represents  an  hyperbola  with  the 
foci  on  OX.  For  if  we  place  —  b2  =  a2(l  —  e2),  we  find  for 
the   eccentricity   of  the   hyperbola 


Va2+b2 
e  = ■ 


and  may  find  the  position  of  the  foci  from  the  equations 
OF  =  -OF'=  ae. 


Similarly,  any  equation  of  the  form 


a*^b2 

represents  an  hyperbola  with  the  foci  on  OY. 
If  the  two  hyperbolas, 

*:-|S  =  l     and     -2,  +  g-l 

a2      b2  a2      b1 

have  the  same  values  for  a  and  b,  each  is  said  to  be  the  conjugate 
hyperbola  to  the  other. 

If  b  =  a,  the  hyperbola  is  called  an  equilateral  hyperbola,  and  its 
equation  is  either  x2—  y2  =  a2  or  —  x2  +  y2  =  a2. 

Finally,  it  is  evident  that  either 

(x-h)2     (y-k)2 
a2  b2 

a2  b" 

is  the  equation  of  an  hyperbola  with  its  center  at  the  point  (li,  Tc). 
44.  The  parabola.  A  parabola  is  the  locus  of  a  point  equally 
distant  from  a  fixed  point  and  a  fixed  straight  line.  The  fixed 
point  is  called  the  focus  and  the  fixed  straight  line  the  directrix. 
Let  the  line  through  the  focus  perpendicular  to  the  directrix 


THE  PARABOLA 


81 


be  taken  as  the  axis  of  x,  and  let  the  origin  be  taken  on  this 

line,  halfway  between  the  focus  and  the  directrix.    Let  us  denote 

the  abscissa  of  the  focus  by  p.    In  fig.  76  let  F  be  the  focus,  RS 

the  directrix  intersecting  OX  at  D,  and  P 

any  point  on  the  curve.    Then  F  is  Q>,  0), 

D  is  (—  p,  0),  and  the  equation  of  ES  is 

x  =  —  p.     Draw  from  P  a  line  parallel  to 

OX,  intersecting  ES  in  X.     If  F  is  on  the 

right  of  ES,  P  must  also  lie  on  the  right 

of  ES,  and,  by  the  definition, 


SY 

V           p^^ 

a 

D 

\F 

FP  =  NP. 


Fig.  76 


If,  on  the  other  hand,  F  is  on  the  left  of  ES,  P  is  also  on  the 
left  of  ES,  and 


In  either  case 
But,  by  §  5, 

and 

hence 

which  reduces  to 


FP=PX=-M'. 
FP*=NP\ 

FP'2  =  (x-pY-hf, 
NP=x+p-, 

(x-py  +  f  =  (x+Py, 

y2=4ju: 


(1) 


Any  point  on  the  parabola  then  satisfies  equation  (1). 
Conversely,  it  is  easy  to  show  that  if  a  point  satisfies  (1),  it 
lies  so  that  FP=±XP,  and  hence  lies  on   the  parabola. 

Equation  (1)  shows  that  the  curve  is  Symmetrical  with 
respect  to  OX,  that  x  must  have  the  same  sign  as  p,  and  that  // 
increases  as  x  increases  numerically.  The  position  of  the  curve 
is  as  shown  in  fig.  76  when  p  is  positive.  When  p  is  negative,  F 
lies  at  the  left  of  0,  and  the  curve  extends  toward  the  negative 
end  of  the  axis  of  x. 

Similarly,  the  equation  x2  =  4  py  represents  a  parabola  for 
which  the  focus  lies  on  the  axis  of  y,  and  which  extends 
toward  the  positive  or  the  negative  end  of  the  axis  of  y 
according  as  p  is  positive  or  negative.  In  all  cases  0  is 
called  the  vertex  of  the  parabola,  and  the  line  determined 
by   O  and  F  is   called  its   axis. 


82 


CERTAIN  CURVES 


A  more  general  equation  of  the  parabola  is  evidently 
(y  —  ky=  kp(x  —  h) 
or  Qx  —  hy  =  4  p  (y  —  &), 

the  vertex  in  either  case  being  at  the  point  (A,  F).  The  work  of 
locating  the  parabola  in  the  plane  is  illustrated  in  the  following 
example. 

Ex.  Show  that  y2  +  y  —  3  x  +  1  =  0  is  a  parabola,  and  locate  it  in 
the  plane. 

The  equation  may  be  written 

?y2  +  y  =  3  x  -  1, 
or  if  +  y  +  \  =  3  x  —  1  +  \, 

which  reduces  to  (//  +  \)"  =  3  (.r  —  \). 

Hence  the  vertex  is  at  the  point  (\,  —  J);  the  equation  of  the  axis  is 
y  +  1-  =  0,  or  2  y  +  1  =  0  ;  the  focus  is  at  the  point  (I  +  4,  —  I),  or  (1,  —  J-) ; 
and  the  equation  of  the  directrix  is  x  —  \  =  —  J,  or  2  x  +  1  =  0. 

45.  If  IKx^  y^)  and  i^(#2,  y.^)  are  two  points  on  the  parabola 
y=4^(fig.  77),  then 

y"  =  4JFi« 

y22  =  4^2; 

(1) 


whence 

yl 

yl 

which  may  be  wi 

itten 

(?y.y 

(2*/2)2 

(2) 


Fig.   77 


if  both   numerator  and   denominator  of  the  left-hand  fraction 
are  multiplied  by  4. 

From  the  symmetry  of  the  parabola,  2  yx=  Q^  and  Q2P2  =  2  ?/2; 
OM  ,  (2)  becomes 


and  since  xx=  OMl  and  x, 


Q^i       OM1 


Q,P, 


OIL 


(3) 


That  is,  the  squares  of  any  two  chords  of  a  parabola  tvhich  are 
perpendicndar  to  the  axis  of  the  parabola  are  to  each  other  as  their 
distances  from  the  vertex  of  the  parabola. 


THE  CONIC 


The  figure  bounded  by  the  parabola  and  a  chord  perpendic- 
ular to  the  axis  of  the  parabola,  as  QlOPl  (fig.  77),  is  called  a 
parabolic  segment.  The  chord  is  called  the  base  of  the  segment, 
the  vertex  of  the  parabola  is  called  the  vertex  of  the  segment, 
and  the  distance  from  the  vertex  to  the  base  is  called  the 
altitude  of  the  segment. 

46.  The  conic.  A  conic  is  the  locus  of  a  point  the  distance  of  which 
from  a  fixed  point  is  in  a  constant  ratio  to  its  distance  from  a  fixed 
straight  line. 

The  fixed  point  is  called  the  focus, 
the  fixed  line  the  directrix,  and  the 
constant  ratio  the  eccentricity. 

We  shall  take  tin;  directrix  as  the 
axis  of  y  (fig.  78),  and  a  line  through 
the  focus  F  as  the  axis  of  x,  and 
shall  call  the  focus  (<■,  0),  where  c 
represents  OF  and  is  positive  or 
negative  according  as  F  lies  to  the 
right  or  the  left  of  0. 

Let  P  be  any  point  on  the  conic; 
connect  P  and  /\  and  draw  I'X  per- 
pendicular to  0  Y.  Then,  by  definition, 

FP  =  ±e.NP,  (1) 


J 

2f 

r 

/ 
/ 

,( 

0 

\       & 

\ 
\ 
\ 

\ 
\ 
\ 
\ 
\ 
\ 
\ 
\ 

Fig.  7k 


according  as/'  is  on  the  right  or  the  left  of  OY.    In 
FP2=et.NPi. 


both 


But  FP2=(x-cY+g 
any  point  on  the  conic 


2    by  £  5,  and  NP  =  x.    Therefore  for 

<yf+<f=e\r. 


(2) 


It  is  easy  to  show,  conversely,  that  if  the  coordinates  of  P  sat- 
isfy (2),  P  satisfies  (1).    Hence  (2)  is  the  equation  of  the  conic. 

It  is  clear  that  the  parabola  is  a  special  case  of  a  conic,  for  the 
definition  of  the  latter  becomes  that  of  the  former  when  e  =  1. 

It  is  also  not  difficult  to  show  that  the  ellipse  is  a  special  case 
of  a  conic,  where  the  eccentricity  is  e  of  §  38  and  <  1. 


84 


(KIM1  AIN  CURVES 


r  If 

For  if  P  (fig.  79)  is  a  point  on  the  ellipse  "--  +  ~-  =  1,  we  found 
in  §  38  that  a 


FP  =  a-  ex, 


F'P  =  a  +  ex; 


F'P 


+  x 


1) 


If  now  we  take  the  point 

so  that  0D  =  ~,   and  D' 

e 


so  that  OD'  =  —  -,  and  if  we 

e 

draw  the  lines  DS  and  D'S' 
perpendicular  to  OX,  the 
line  N'PN  perpendicular  to 
JDS,  and  the  ordinate  MP, 
we  have 


Fig.  79 


Therefore 


--X  =  0D-  OM=MD  =  PN, 

e 

-  +  x=I)'0+OM=D'M=N'P. 

e 

FP  =  e-  PN,  F'P  =  e  •  N'P. 


The  ellipse  has,  therefore,  two  directrices  at  the  distances  ±  - 

e 

from  the  center.     When  the  ellipse  is  a  circle,  e  =  0  and  the 
directrices  are  at  infinity. 

In  a  similar  manner  we  may  show  that  the  hyperbola  is  a 
special  case  of  a  conic  where  e  >  1. 

47.  The  witch.    Let  OB  A  (fig.  80)  be  a  circle,  OA  a  diameter, 
and   LK  the    tangent    to  y 

the  circle  at  A.  From  0 
draw  any  line  intersect- 
ing the  circle  at  B  and 
LK  at  C.  From  B  draw 
a  line  parallel  to  LK  and 
from  C  a  line  perpendic- 
ular to  LLC,  and  call  the  intersection  of  these  two  lines  P. 
The  locus  of  P  is  a  curve  called  the  witch. 


Fig.  80 


THE  CISSOID 


85 


To  obtain  its  equation  we  will  take  the  origin  at  0  and  the 
line  OA  as  the  axis  of  y.  We  will  call  the  length  of  the  diameter 
of  the  circle  2  a.  Then,  by  continuing  CP  until  it  meets  OX  at 
M  and  calling  (x,  y)  the  coordinates  of  P,  we  have 

0  31  =  x,         MP  =  y,  OA  =  MC  =  2  a. 


In  the  triangle  03IC, 


MP 
MC 


OB 
OC 


OB  .  PC 

~oc2~' 


(1) 


Draw  AB.    Then  OBA  is  a  right  angle,  and  consequently 


OB  .00  =  OA2;      also  0 C*  =  03f+  MC*. 


Therefore 


that  is, 


and  iinally, 


MP 
MC 

JL 
2a 


OA 


OW+MC 
4  a2 


x-  +  4  <i- 
8  az 


*      a»+4ta* 
Solving  (4)  for  x,  we  have 


;(2) 


(3) 


(■*) 


x  =  ±  2  a 


\ 


y 


This  shows  that  the  curve  is  sym- 
metrical with  respect  to  OY,  that  y 
cannot  be  negative  nor  greater  than 
2  a,  and  that  //  =  0  is  an  asymptote. 

48.  The  cissoid.  Let  01) A  (fig.  81)  . 
be  a  circle  with  the  diameter  OA, 
and  let  LK  be  the  tangent  to  the 
circle  at  A.  Through  0  draw  any 
line  intersecting  the  circle  in  D  and 
LK  in  E.  On  OE  lay  off  a  distance 
OP,  equal  to  DE.  Then  the  locus  of 
P  is  a  curve  called  the  cissoid. 

To  find  its  equation  we  will  take  O  as  the  origin  of  coordi- 
nates and  OA  as  the  axis  of  x,  and  will  call  the  diameter  of  the 
circle  2  a.    Join  A  and  D  and  draw  MP  perpendicular  to  OA. 


Fig.  81 


86 


CE11TAIN  OUJiVES 


Denoting  angle  MOP  by  0,  we  have 
OJ0=2aaec0, 
OD=2acos0; 
DE  =  OE  -0D=2a  (sec  0  -  cos  0). 
OP=2a(sec0-cos0). 
x  =  OM=  OP  cos0 
=  2a(l-cos26>) 
=  2asin20. 
MP  v 


whence 
Therefore 
Now 


But 


sin  0  = 


Substituting  in  (5),  we  have 


whence 

2  a 

This  equation  is  satisfied  by  the 
coordinates  of  any  point  upon  the 
cissoid.    It  may  be  written 


OP 

y/a? 

+  tf 

2  ay2 

x2  +  f 

xs 

=±x4n? 


x 


From  this  it  appears  that  the 
curve  is  symmetrical  with  respect 
to  OX,  that  no  value  of  x  can  be 
greater  than  2  a  or  less  than  0, 
and  that  the  line  x  =  2  a  is  an 
asymptote. 

49.  The  strophoid.  Let  LK  and 
ES  (fig.  82)  be  two  straight  lines 
intersecting  at  right  angles  at  0, 
and  let  A  be  a  fixed  point  on  LK. 
Through  A  draw  any  straight  line 
intersecting  ES  in  D,  and  lay  off 
on  AD  in  either  direction  a  distance  DP  equal  to  OD. 
locus  of  P  is  a  curve  called  the  strophoid. 


(1) 
(2) 
(3) 
(4) 


(5) 
(6) 

(7) 

(8) 


The 


THE  STHOPHOID  87 

To  find  its  equation  take  LK  as  the  axis  of  x  and  RS  as  the 
axis  of  ?/,  and  let  OA  =  a.  By  the  definition,  the  point  P  may 
fall  in  any  one  of  the  four  quadrants.  If  we  take  the  positive 
direction  on  AD  as  measured  from  A  towards  D,  we  have 

OD  =  PD 
when  P  is  in  the  first  quadrant, 

OD=-PD 
when  P  is  in  the  second  quadrant, 

when  P  is  in  the  third  quadrant,  and 

-  OD  =  I'D 

when  P  is  in  the  fourth  quadrant. 

These  four  equations  are  equivalent  to  the  single  equation 


o!r  = 

■FiA 

(1) 

Draw  PM 

parallel  to  OT  and  denote  the 

ai 

igle 

J/J/' 

by 

0. 

Then 

0D  = 

:  a  tan  0 

(2) 

and 

PD  = 

./Sec  0. 

(3) 

Substitutii 

ig  in  (1), 

we  have 
a2 tan2  0  = 

a?  Bee*  6. 

(4) 

But 

tan  0  = 

MP        y 

(5) 

MA     a- 

X 

Substituting  in  (4),  we  have 

which  may  be  reduced  to         y  =  ±  x-y\—  —  •  (7) 

This  shows  that  the  curve  is  symmetrical  with  respect  to  OX, 
that  no  value  of  x  can  be  less  than  —  a  or  greater  than  +  a, 
and  that  x  =  —  a  is  an  asymptote. 


88 


CERTAIN  CURVES 


50.  Use  of  the  equation  of  a  curve.  The  use  of  the  equation 
of  a  curve  in  solving  geometrical  problems  is  illustrated  in 
the  following   problems : 

Ex.  1.   Prove  that   in   the 

ellipse  the  squares  of  the 
ordinates  of  any  two  points 
are  to  each  other  as  the 
products  of  the  segments  of 
the  major  axis  made  by  the 
feet  of  these  ordinates. 

We  are  to  prove  that  (fig.  S3) 

MJ\2  _  A'MX  ■  MXA 

Mjn~  A'M2.  M^A 

Let  Px  be  (xv  yx)  and  let 
P2  be  (x2,  y2).    Then 

■,,2 

whence 

But  yx  =  M1PV  a  +  xx  =  A'O  +  OM1  =  A'MV  a-x1  =  OA  - OMx  =  MXA, 
y2  —  M2P2,  a  +  x2  =  A'M2,  a  —  x2  =  M2A.    Hence  the  proposition  is  proved. 

Ex.  2.  If  MlPl  is  the  ordinate  of  a 
point  Pt  of  the  parabola  y2  =  <±px,  and 
a  straight  line  drawn  through  the  middle 
point  of  M1P1  parallel  to  the  axis  of  x 
cuts  the  curve  at  Q,  prove  that  the  inter- 
cept of  the  line  MM  on  the  axis  of  y 
equals   %M1PV 

Let  Px  (fig.  84)    be    (xv  yx).     Then 

y?  * 

x.  =  — — ,     from     the     equation     of     the 
ip 

parabola. 

By   construction,    the    ordinate    of    Q 


F 

y^^ 

7^ 

0 

I 

f. 

Vi 


Since   Q  is   on   the   parabola,  its 
is  found  by  placing  y  =  ^  in  if-  =  4  px.   Then  Q  is 


Fig.  84 


\l6p    2/' 


and 
\Wp    H/' 

M  is  (xv  0),  which  is  the  same  as  (-p->  0  ).    Hence  the  equation  of  MM 

is,  by  §  30,  ^P      I 

8^  +  3^-2^  =  0. 

The  intercept  of  this  line  on  OFis  §  yx  =  ^MlPv  which  was  to  be  proved. 


EMPIRICAL  EQUATIONS  89 

51.  Empirical  equations.  We  have  met  in  §  8  examples  of 
related  quantities  for  which  pairs  of  corresponding  values  have 
been  found  by  experiment,  but  for  which  the  functional  relation 
connecting  the  quantities  is  not  known.  In  such  a  case  it  is  often 
desirable  to  find  an  equation  which  will  represent  this  relation, 
at  least  approximately.  The  method,  in  general,  is  to  plot  the 
points  as  in  §  8  and  then  fit  a  curve  to  them.  At  best  this  work 
is  approximate,  the  result  depending  largely  on  the  judgment  of 
the  worker,  and  in  complicated  cases  it  demands  methods  too 
advanced  for  this  book.  We  shall  discuss  a  few  simple  exam- 
ples, to  illustrate  merely  the  fundamental  principles  involved. 

The  simplest  case  is  that  in  which  the  plotted  points  appear 
to  lie  on  a  straight  line,  or  nearly  so.  If  the  two  related  quan- 
tities are  x  and  y,  the  relation  between  them  is  expressed  by  the 
equation  y=mx  +  b,  (1) 

where  m  and  b  are  to  be  determined  to  fit  the. data.    In  practice 

the  points  are  plotted,  and  it  appears  that  a  straight  line  may  be 

so  drawn  that  the  points  either  lie  on  it  or  are  close  to  it  and 

about  evenly  distributed  on  both  sides  of  it.    The  straight  line 

having  been  drawn,  its  equation  may  be  found  by  means  of  two 

points  on  it,  which  may  be  either  two  of  the  original  data  or  any 

two  points  of  the  graph.    This  method  is  illustrated  in  Ex.  1. 

Closely  connected  with  this  case  are  two  others.    Suppose  the 

relation  between  the  two  quantities  .'■  ami  y  is  known  or  assumed 

to  be  of  the  form  „  .ON 

V  =  "•'  (2) 

or  y  =  a^xi  (&) 

where  a,  b,  and  n  are  to  be  determined  to  fit  the  given  numerical 

values  of  x  and  y.    By  taking  the  logarithms  of  both  sides  of 

these  equations,  we  have  respectively 

log  y  =  n  log  x  +  log  a  (4) 

and  log  y  =  (log  b~) x  +  log  a ;  (5) 

or,  if  we  place  log  y  =  y',  log  x  =  x',  log  a  =  b',  log  b  =  m, 

y'  =  nx'+b',  (G) 

y'  =  mx  +  V.  (7) 


<)0 


CERTAIN  CURVES 


We  may  now  plot  the  points  (V,  y'~)  or  (x,  y'~)  and  determine 
the  straight  line  on  which  they  lie  approximately.  The  equations 
(0)  and  (7)  having  thus  been  found,  the  return  to  equations 
(2)  and  (3)  is  easy.    This  method  is  illustrated  in  Ex.  2. 

When  the  use  of  a  straight  line  either  directly  or  by  aid  of 
logarithms  fails,  the  attempt  may  be  made  to  fit  a  parabola 

y  =  a  +  bx  +  ex2  (9) 

to  the  points  of  the  plot.  Since  three  points  are  sufficient  to  de- 
termine the  constants  of  the  equation,  the  parabola  may  be  made 
to  pass  through  any  three  of  the  plotted  points.  This  parabola 
may  then  be  tested  to  see  if  it  passes  reasonably  near  to  the 
other  points.  This  method  is  illustrated  in  Ex.  3. 
Other  curves  with  equations  of  the  form 

y  =  a  +  bx  +  cx2  +  dx*-\ +  lxn 

may  also  be  used.  In  this  case  the  number  of  points  through 
which  the  curve  may  be  exactly  drawn  is  equal  to  the  number 
of  arbitrary  coefficients. 

In  all  these  cases  it  is  often  convenient  to  use  different  scales 
for  x  and  y,  the  proper  allowance  being  made  in  the  calculations. 
This  is  illustrated  in  Ex.  2. 

Ex.  1.  Corresponding  values  of  two  related  quantities  x  and  y  are 
given  by  the  following  table  : 


X 

1 

2 

4 

G 

10 

V 

1.8 

2.2 

2.9- 

3.9 

6.1 

Find    the     empirical     equation 
connecting    them. 

We  plot  the  points  (x,  y) 
and  draw  the  straight  line  as 
shown  in  fig.  85.  The  straight 
line  is  seen  to  pass  through  the 
points  (0,  1)  and  (2,  2).  Its 
equation  is  therefore 

y  =  .5  x  +  1, 

which  is  the  required  equation. 


] 

T 

L, 

Si 

0 

1 

Fig.  85 


EMPIKICAL  EQUATIONS 


91 


Ex.  2.  Corresponding  values  of  pressure  and  volume  taken  from  an 
indicator  card  of  an  air  compressor  are  as  follows  : 


p 

18 

21 

20.5 

33.5            44             62 

V 

.635 

.556 

.475 

.397          .321          .243 

Find  the  relation  between  them  in  the  form  pvn  —  c. 

Writing  the  assumed  relation  in  the  form  p  =  cv~*  and  taking  the 
logarithms  of  both  sides  of  the  equation,  we  have 

logp  =  -  nlogw  +  loge, 
or  y  =  —  nx  +  b, 

where  y  =  logp,  x  =  log  v,  and  b  =  log  c. 

The  corresponding  values  of  x  and  y  arc 


X  =  1 1  tg  V 

-.1972 

-  .2549 

-  .3233 

-  .41112 

-.4935       -.6144 

y  =  logp 

1.2553 

1.3222 

1.4232 

1.5250 

1.0435    j      1.7924 

We  assume  on  the  ar-axis  a  scale  twice  as  large  as  that  on  the  //-axi 

the  points  (.r,  ?/),  and  draw  the  straight  line  as  shown  in  fig.  86 

construction    should    be    made 

on    large-scale    plotting   paper. 

The  line  is  seen  to  pass  through 

the    points    (—.05,   1.075)   and 

(—.46,   1.6).     Its    equation   is 

therefore 


:s: 


x 


\ 


'.BO-Jtf-JiD-SS-aO-.SB-.SO-lS—lO-  0B  o 
Fiu.  86 


y  =  — 1.28  x+  1.01. 

Hence  n  =  1.28,  log  c  =  1.01, 
c  =  10.2,  and  the  required  re- 
lation between  p  and  V  is 

pv1-2*  -  10.2. 

Ex.  3.    Corresponding   values   of   two   related   cpuantities   x  and   y   are 
given  by  the  following  table : 


X                  1 

2 

3 

4 

5 

y       1.37 

.08 

.41 

.54 

1.05 

Find  the  empirical  equation  connecting  them. 


92 


CERTAIN  CURVES 


If  we  plot  the  points 
we  assume 


is  in  fig.  87,  they  suggest  a  parabola.   Accordingly 
w  =  a  +  hx  +  ex2 


and  determine  a,  h,  and  c,  so  that  the  curve  will  pass  through  the  first, 
third,  and  last  points.    The  equations  for  a,  b,  and  c  are 


1.37 

=  a  +  b  +  c, 

.11 

=  a  +  3  b  +  9  r, 

1.05 

=  a  +  5  h  +  25  c ; 

w 

H'UI'I 

«  =  2 

.45,  &  =  -  1.28,  and 

c 

=  .2. 

The 

required    equation 

is 

therefore 

r 

l 

O 

l 

2.45  -  1.28  x  +  .2  x2 


Fig.  8< 


If  we  substitute  for  x  in  this  equation  the  values  2  and  4,  we  find  the 
corresponding  values  of  y  to  he  .69  and  .53.  This  shows  that  the  curve  passes 
reasonably  near  to  the  points  of  the  plot  which  were  not  used  in  computing 
the  coefficients. 

PROBLEMS 

1.  Find  the  equations  of  the  locus  of  a  point  the  distance  of  which 
from  the  axis  of  x  equals  five  times  its  distance  from  the  axis  of  y. 

2.  Find  the  equations  of  the  locus  of  a  point  the  distance  of  which 
from  the  axis  of  a;  is  3  more  than  twice  its  distance  from  the  axis  of  y. 

3.  Find  the  equation  of  the  locus  of  a  point  the  distances  of  which 
from  (2,  —  1)  and  (—  3,  2)  are  equal. 

4.  Find  the  equations  of  the  locus  of  a  point  equally  distant 
from  the  lines  3  ,r  —  5  y  —  15  =  0  and  5x-  3y  +  l  =  0. 

5.  Find  the  equations  of  the  bisectors  of  the  angles  between  the 
lines  9ir  +  2?/  —  3  =  0  and  1  x  —  6 y  +  2  =  0. 

6.  Find  the  equations  of  the  bisectors  of  the  angles  between  the 
lines  3.x  +  4y— 7=0  and  12 x  —  5 y  + 1  =  0. 

7.  A  point  moves  so  that  its  distance  from  the  axis  of  y  equals 
its  distance  from  the  point  (5,  0).    Find  the  equation  of  its  locus. 

8.  Find  the  equation  of  the  locus  of  a  point  the  distance  of 
which  from  the  axis  of  x  is  one  half  its  distance  from  (0,  2). 

9.  A  point  moves  so  that  the  square  of  its  distance  from  the 
point  (0,  3)  equals  the  cube  of  its  distance  from  the  axis  of  y.  Find 
the  equation  of  its  locus. 


PROBLEMS  93 

10.  Find  the  equation  of  the  locus  of  a  point  the  distance  of  which 
from  the  line  x  =  3  is  equal  to  its  distance  from  (4,  —  2) . 

11.  Find  the  equation  of  the  locus  of  a  point  which  moves  so  that 
the  slope  of  the  straight  line  joining  it  to  (a,  a)  is  one  greater  than 
the  slope  of  the  straight  line  joining  it  to  the  origin. 

12.  A  point  moves  so  that  its  distance  from  the  origin  is  always 
equal  to  the  slope  of  the  straight  line  joining  it  to  the  origin.  Find 
the  equation  of  its  locus. 

13.  Find  the  equation  of  the  locus  of  a  point  the  distance  of  which 
from  the  line  3  a?  +  4  y  —  6  =  0  is  twice  its  distance  from  (2, 1). 

14.  Find  the  equation  of  the  circle  having  the  center  (3,  —  5)  and 
the  radius  4. 

15.  Find  the  equation  of  the  circle  having  the  center  i  —  |}  g  |  and 

the  radius  2. 

16.  Find  the  points  at  which  the  axis  of  x  intersects  the  circle 
having  as  diameter  the  straight  line  joining  (1,  2)  and  (—  3,  —  4). 

17.  Find  the  equation  of  the  circle  having  as  diameter  that  part 
of  the  line  Sx  —  4//  +  12  =  0  which  is  included  between  the  coor- 
dinate axes. 

18.  Find  the  equation  of  the  circle  having  as  diameter  the  common 
chord  of  the  two  circles 

x*  +  y*+±x  —  4y  —  2  =  0   and   x-  +  f  -  2 x  +  2 y  -  14  =  0. 

19.  Find  the  ('([nations  of  the  circles  of  radius  a  winch  are  tangent 
to  the  axis  of  y  at  the  origin. 

20.  Find  the  center  and  the  radius  of  the  circle 

./■-  +  //-  +  26  x  +  1 6  //  -  42  =  0. 

21.  Find  the  center  and  the  radius  of  the  circle 

2  x*  +  2  y2  -r-  f>  .r  +  3  y  -10  =  0. 

22.  Find  the  equation  of  a  straight  line  passing  through  the  cen- 
ter of  the  circle  x'2  +  if  —  4.r  -f-  2y  —  5  =  0  and  perpendicular  to 
the  line  x  —  2 //  +  1  =  0.    How  near  the  origin  does  the  line  pass  '.' 

23.  Prove  that  two  circles  are  concentric  if  their  equations  differ 
only  in  the  absolute  term. 


94  CERTAIN  CURVES 

24.   Show   that    the    circles    x2  +  if  +  2  Gx  +  2Fy  +  C  =  0   and 
a,2+Z/'2+  2r;'.r  +  2F'y  +  C  =  0  are  tangent  to  each  other  if 


V(G-6")2+(F-JP')2  =  Vg2  +  J"2  -  C  ±  V<?'2  +  Fn  -  C". 

25.  Find  the  equation  of  the  circle  which  passes  through  the 
points  (0,  2),  (2,  0),  and  (0,  0). 

26.  Find  the  equation  of  the  circle  circumscribing  the  triangle 
with  the  vertices  (0, 1),  (-  2,  0),  and  (0,  -  1). 

27.  Find  the  equation  of  the  circle  circumscribing  the  isosceles 
triangle  of  which  the  altitude  is  5  and  the  base  is  the  line  joining 
the  points  (—  4,  0)  and  (4,  0). 

28.  Find  the  equation  of  the  circle  circumscribed  about  the  tri- 
angle the  sides  of  which  are  x  +  2  y  —  3  =  0,  3  cc  —  y  —  2  =  0,  and 
2x-3y-6  =  0. 

29.  Find  the  equation  of  the  circle  passing  through  the  point 
(—3,  4)  and  concentric  with  the  circle  x2  +  y2  +  3  x  —  4  y  —  1  =  0. 

30.  A  circle  which  is  tangent  to  both  coordinate  axes  passes 
through  (4,  —  2).    Find  its  equation. 

31.  The  center  of  a  circle  which  is  tangent  to  the  axes  of  x  and 
of  y  is  on  the  line  2>x  —  5  y  4- 15  =  0.    What  is  its  equation  ? 

32.  A  circle  of  radius  5  passes  through  the  points  (4,  —  2)  and 
(5,  —  3).    What  is  its  equation  ? 

33.  The  center  of  a  circle  which  passes  through  the  points  (—  2,  4) 
and  (—  1,  3)  is  on  the  line  2x  —  3y  +  2  =  0.    What  is  its  equation  ? 

34.  A  circle  which  is  tangent  to  OX  passes  through  (— 1,  2)  and 
(6,  9).    What  is  its  equation  ? 

35.  The  center  of  a  circle  which  is  tangent  to  the  two  parallel 
lines  x  —  2  =  0  and  x  —  6  =  0  is  on  the  line  y  =  3  x  —  6.  What  is 
its  equation  ? 

36.  The  center  of  a  circle  is  on  the  line  2x  +  y  4-  3  =  0.  The 
circle  passes  through  the  point  (3,  1)  and  is  tangent  to  the  line 
4a:  —  3y  — 14  =  0.    What  is  its  equation  ? 

37.  The  center  of  a  circle  is  on  the  line  x  4-  2y  — 10  =  0  and  the 
circle  is  tangent  to  the  two  lines  2x  —  3y+9  =  0  and  3x  —  2y  +  1  =  0. 
What  is  its  equation  ? 

38.  Given  the  ellipse  9  x2  +  25  y2  =  225,  find  its  semiaxes, 
eccentricity,  and  foci. 


•    PROBLEMS  95 

39.  Given  the  ellipse  3  a-2  +  4  y2  =  2,  find  its  semiaxes,  eccen- 
tricity, and  foci. 

40.  Find  the  vertices,  eccentricity,  and  foci  of  the  ellipse 
4  x2  +  2  if  =  1. 

41.  Find  the  center,  vertices,  eccentricity,  and  foci  of  the  ellipse 
4  a-2  +  9  if  +  16 x  -18  y  -11  =  0. 

42.  Find  the  center,  vertices,  eccentricity,  and  foci  of  the  ellipse 
16  x2  +  9  if  - 16  x  +  6  y  - 139  =  0. 

43.  Find  the  equation  of  th»  ellipse  when  the  origin  is  at  the 
left-hand  vertex  and  the  major  axis  lies  along  OX. 

44.  Find  the  equation  of  the  ellipse  when  the  origin  is  taken  at 
the  lower  extremity  of  the  minor  axis  and  the  minor  axis  lies 
along  OY. 

x2       f 

45.  Determine  the  semiaxes  a  and  b  in  the  ellipse  -j  -f-  j-  =  1   so 

that  it  shall  pass  through  (2,  3)  and  (—1,  —  4). 

46.  Find  the  equation  of  an  ellipse  if  its  axes  are  8  and  4,  its 
center  is  at  (2,  —  3),  and  its  major  axis  is  parallel  to  OX. 

47.  Find  the  equation  of  an  ellipse  if  its  axes  are  §  and  £,  its 
center  is  at  (1,  —1),  and  its  major  axis  is  parallel  to  OY. 

48.  If  the  vertices  of  an  ellipse  are  (±  6,  0)  and  its  foci  are 
(±4,  0),  find  its  equation. 

49.  Find  the  equation  of  an  ellipse  when  the  vertices  are  (±  4,  0) 
and  one  focus  is  (2,  0). 

50.  Find  the  equation  of  an  ellipse  when  the  vertices  are  (0,  2) 
and  (0,  —  4)  and  one  focus  is  at  the  origin. 

51.  Find  the  equation  of  the  ellipse  the  foci  of  which  are  (±  4,  0) 
and  the  major  axis  of  which  is  10. 

52.  Find  the  equation  of  the  ellipse  the  foci  of  which  are  (0,  ±  3) 
and  the  major  axis  of  which  is  12. 

53.  Find  the  equation  of  an  ellipse  when  its  center  is  at  the 
origin,  one  focus  is  at  the  point  (—  4,  0),  and  the  minor  axis  is 
equal  to  6. 

54.  Find  the  equation  of  the  ellipse  the  foci  of  which  are  (1,  ±  2) 
and  the  major  axis  of  which  is  6. 

55.  Find  the  equation  of  an  ellipse  the  eccentricity  of  which  is  § 
and  the  foci  of  which  are  (0,  ±  5). 


96  CERTAIN  CURVES 

56.  The  center  of  an  ellipse  is  at  the  origin  and  its  major  axis 
lies  on  OX.  If  its  major  axis  is  6  and  its  eccentricity  is  i,  find  its 
.equation. 

57.  The  center  of  an  ellipse  is  at  (—  2,  3),  and  its  major  axis  is 
parallel  to  OF  and  8  units  in  length.  Its  eccentricity  is  ^.  Find  its 
equation. 

58.  The  center  of  an  ellipse  is  at  (1,  2),  its  eccentricity  is  \,  and 
the  length  of  its  major  axis,  which  is  parallel  to  OY,  is  8.  What  is 
the  equation  of  the  ellipse  ? 

59.  Find  the  equation  of  an  ellipse  the  eccentricity  of  which  is  ^ 
and  the  ordinate  at  the  focus  is  4,  the  center  being  at  the  origin  and 
the  major  axis  lying  on  OX. 

60.  Find  the  eccentricity  and  the  equation  of  an  ellipse  if  the 
foci  lie  halfway  between  the  center  and  the  vertices,  the  center 
being  at  the  origin  and  the  major  axis  lying  on   OX. 

61.  Find  the  equation  and  the  eccentricity  of  an  ellipse  if  the 
ordinate  at  the  focus  is  one  third  the  minor  axis,  the  center  being 
at  the  origin  and  the  major  axis  lying  on  OX. 

62.  Find  the  eccentricity  of  an  ellipse  if  the  straight  line  connect- 
ing the  positive  ends  of  the  axes  is  parallel  to  the  straight  line  joining 
the  center  to  the  upper  end  of  the  ordinate  at  the  left-hand  focus. 

63.  Given  the  hyperbola  —  —  £-  =  1.  find  its  eccentricity,  foci, 
and  asymptotes. 

64.  Given  the  hyperbola  4  a-2  —  9  if  —  36,  find  its  eccentricity, 
foci,  and  asymptotes. 

65.  Find  the  center,  eccentricity,  foci,  and  asymptotes  of  the 
hyperbola  9  r2  -  4  f  -  36  x  -  24  y  -  30  =  0. 

66.  Find  the  center,  eccentricity,  foci,  and  asymptotes  of  the 
hyperbola  2  x2  -  3  f  +  4  x  +  12  y  +  4  =  0. 

67.  Find  the  equation  of  an  hyperbola  if  its  transverse  axis  is 
V3,  its  conjugate  axis  Vf,  its  center  at  (1,  —  2),  and  its  transverse 
axis  parallel  to  OX. 

68.  Find  the  equation  of  an  hyperbola  if  its  transverse  axis  is 
5,  its  conjugate  axis  3,  its  center  (—  2,  3),  and  its  transverse  axis 
parallel  to  OY. 

69.  Find  the  equation  of  the  hyperbola  when  the  origin  is  at  the 
left-hand  vertex,  the  transverse  axis  lying  on  OX. 


PROBLEMS  97 

70.  Find  the  equation  of  an  hyperbola  if  the  foci  are  (±  4,  0)  and 
the  transverse  axis  is  6. 

71.  Find  the  equation  of  an  hyperbola  if  the  foci  are  (0,  ±  3)  and 
the  transverse  axis  is  4. 

72.  An  hyperbola  has  its  center  at  (1,  2)  and  its  transverse  axis 
is  parallel  to  OX.  If  its  eccentricity  is  §  and  its  transverse  axis  is 
5,  find  its  equation. 

73.  Find  the  equation  of  an  hyperbola  when  the  vertices  are  (7, 1) 
and  (—  1,  1)  and  the  eccentricity  is  |. 

74.  Find  the  equation  of  an  hyperbola  the  vertices  of  which  are 
halfway  between  the  center  and  the  foci,  the  center  being  at  0  and 
the  transverse  axis  lying  on  OX. 

76.  Find  the  equation  of  the  hyperbola  which  has  the  lines 
y  =  ±  |  x  for  its  asymptotes  and  the  points  (±2,  0)  for  its  foci. 

76.  Find  the  equation  of  the  hyperbola  which  has  the  asymptotes 
y  =  ±  |  x  and  passes  through  the  point  (2,  1). 

77.  Find  the  equation  of  an  equilateral  hyperbola  which  passes 
through  (3,  —  1)  and  has  its  axes  on  the  coordinate  axes. 

78.  Show  that  the  eccentricity  of  an  equilateral  hyperbola  is  equal 
to  the  ratio  of  a  diagonal  of  a  square  to  its  side. 

79.  If  the  vertices  of  an  hyperbola  lie  two  thirds  of  the  distances 
from  the  center  to  the  foci,  find  the  angles  between  the  transverse 
axis  and  the  asymptotes. 

80.  Express  the  angle  between  the  asymptotes  in  terms  of  the 
eccentricity  of  the  hyperbola. 

81.  An  ellipse  and  an  hyperbola  have  the  vertices  of  each  at  the 

x2       ir 
foci  of  the  other.    If  the  equation  of  the  ellipse  is  —  +  ~  =  1,  find 

that  of  the  hyperbola.    Find  the  equations  of  the  directrices  of  the 
two  curves. 

.'-  y2 

82.  Show  that  -3 —  +  -h — — ,  =  1,  where   k   is   an   arbitrary 

quantity,  represents  an  ellipse  confocal  to  -5  +  '—  =  1   when   k2<b2, 

and  represents  an  hyperbola  confocal  to  '—,  +  ^  =  1  when  lr>b2  but 
<  a2,  a2  being  greater  than  U1. 

83.  Find  the  vertex,  axis,  focus,  and  directrix  of  the  parabola 

lf  +  ±y-GX  +  T=0. 


98  CERTAIN  CURVES 

84.  Find  the  vertex,  axis,  focus,  and  directrix  of  the  parabola 
4  .r2  +  4  x  4-  3  y  -  2  =  0. 

85.  Determine  p  so  that  the  parabola  y-  =  A])x  shall  pass  through 
the  point  (—  2,  4). 

86.  The  vertex  of  a  parabola  is  at  the  point  (2,  3),  and  the  parab- 
ola passes  through  the  origin  of  coordinates.  Eind  its  equation,  its 
axis  being  parallel  to  OX. 

87.  The  vertex  of  a  parabola  is  at  the  point  (— 1^,  2),  and  the 
parabola  passes  through  the  point  (—  1,  —  1).  Find  its  equation,  its 
axis  being  parallel  to  OY. 

88.  Find  the  equation  of  the  parabola  when  the  origin  is  at  the 
focus  and  the  axis  of  the  parabola  lies  on  OX. 

89.  Find  the  equation  of  the  parabola  when  the  axis  of  the  curve 
and  its  directrix  are  taken  as  the  axes  of  x  and  y  respectively. 

90.  The  vertex  of  a  parabola  is  (3,  2)  and  its  focus  is  (5,  2).  Find 
its  equation. 

91.  The  vertex  of  a  parabola  is  (—  1,  2)  and  its  focus  is  (—  1,  0). 
Find  its  equation. 

92.  Find  the  equation  of  the  parabola  of  which  the  focus  is 
(2,  —  1)  and  the  directrix  is  the  line  y  —  4  =  0. 

93.  The  vertex  of  a  parabola  is  at  the  point  (—  2,  —  5)  and  its 
directrix  is  the  line  x  —  3  =  0.    Find  its  equation. 

94.  The  vertex  of  a  parabola  is  at  (5,  —  2)  and  its  directrix  is  the 
line  ?/  4-  4  =  0.    Find  its  equation. 

95.  The  focus  of  a  parabola  is  at  the  point  (4,  —  1)  and  its  direc- 
trix is  the  line  y  —  x  =  0.  Construct  the  curve  from  its  definition 
and  derive  its  equation.    What  is  the  equation  of  its  axis  ? 

96.  The  altitude  of  a  parabolic  segment  is  8  ft.  and  the  length  of 
its  base  is  14  ft.  A  straight  line  drawn  across  the  segment  perpen- 
dicular to  its  axis  is  7  ft.  long.  How  far  is  it  from  the  vertex  of 
the  segment  ? 

97 .  An  arch  in  the  form  of  a  parabolic  curve,  the  axis  being  verti- 
cal, is  40  ft.  across  the  bottom,  and  the  highest  point  is  12  ft.  above 
the  horizontal.  What  is  the  length  of  a  beam  placed  horizontally 
across  the  arch  3  ft.  from  the  top  ? 


PROBLEMS  99 

98.  The  cable  of  a  suspension  bridge  hangs  in  the  form  of  a 
parabola.  The  roadway,  which  is  horizontal  and  300  ft.  long,  is  sup- 
ported by  vertical  wires  attached  to  the  cable,  the  longest  wire  being 
90  ft.  and  the  shortest  being  20  ft.  Find  the  length  of  a  supporting 
wire  attached  to  the  roadway  50  ft.  from  the  middle. 

99.  Any  section  of  a  given  parabolic  mirror  made  by  a  plane  pass- 
ing through  the  axis  of  the  mirror  is  a  parabolic  segment  of  which 
the  altitude  is  8  in.  and  the  length  of  the  base  is  12  in.  Find  the 
perimeter  of  the  section  of  the  mirror  made  by  a  plane  perpendicular 
to  its  axis  and  6  in.  from  its  vertex. 

100.  Given  the  ellipse  4  x2  -+-  9  y2  =  36,  find  its  foci  and  directrices. 

101.  Given  the  ellipse  5  x2  +  3  f  —  1,  find  its  foci  and  directrices. 

102.  Given  the  hyperbola  5  x2  —  10  y2  =  50,  find  its  foci  and 
directrices. 

103.  Find  the  equation  of  an  ellipse  when  the  foci  are  (±  3,  0) 
and  the  directrices  are  x  =  ±7. 

104.  Find  the  center,  vertices,  foci,  and  directrices  of  the  ellipse 
9x2  +  25  f  +  30 x  +  40 y  -  184  =  0. 

105.  Find  the  center,  vertices,  foci,  and  directrices  of  the  hyper- 
bola 5  x2  -  4  //  +  10  x  +  16  y  -  31  =  0. 

106.  Find  the  equation  of  a  circle  through  the  vertex  and  the 
ends  of  the  double  ordinate  at  the  focus  of  the  parabola  f  =  Apx. 

107.  Find  the  equation  of  the  circle  through  the  vertex,  the 
focus,  and  the  upper  end  of  the  ordinate  at  the  focus  of  the  parab- 
ola f-  8. v  =  0. 

108.  Find  the  equation  of  a  circle  which  passes  through  the 
vertex  and  the  focus  of  the  parabola  f  =  8  x  and  has  its  center 
on  the  line  x  —  y  +  2  =  0. 

109.  Find  the  equation  of  the  locus  of  a  point  which  moves  so 
that  the  slope  of  the  straight  line  joining  it  to  the  focus  of  the 
parabola  x2  =  8  y  is  three  times  the  eccentricity  of  the  ellipse 
16  a:2 +9/ -144  =  0. 

110.  Find  the  equation  of  the  cissoid  when  the  origin  is  at  the 
center  of  the  circle  used  in  its  definition,  the  direction  of  the  axes 
being  as  in  §  48. 

111.  Find  the  equation  of  the  cissoid  when  its  asymptote  is  the 
axis  of  y  and  its  axis  is  the  axis  of  x. 


100  CERTAIN  CURVES 

112.  Find  the  equation  of  the  strophoid  when  the  asymptote  is 
the  axis  of  y,  the  axis  of  x  being  as  in  §  49. 

113.  Find  the  equation  of  the  strophoid  when  the  origin  is  at 
A  (fig.  82),  the  axes  being  parallel  to  those  of  §  49. 

114.  Show  that  the  lines  y  =  ±  x  intersect  the  strophoid  at  the 
origin  only,  and  find  the  equation  of  the  curve  referred  to  these 
lines  as  axes. 

115.  Find  the  equation  of  the  witch  when  LK  (fig.  80)  is  the  axis 
of  x  and  OA  the  axis  of  y. 

116.  Find  the  equation  of  the  witch  when  the  origin  is  taken  at 
the  center  of  the  circle  used  in  constructing  it,  the  axes  being  par- 
allel to  those  of  §  47. 

117.  Show  that  the  locus  of  a  point  which  moves  so  that  the  sum 
of  its  distances  from  two  fixed  straight  lines  is  constant  is  a  straight 
line. 

118.  Find  the  equations  of  the  locus  of  a  point  equally  distant 
from  two  fixed  straight  lines. 

119.  A  point  moves  so  that  its  distances  from  two  fixed  points 
are  in  a  constant  ratio  k.  Show  that  the  locus  is  a  circle  except 
when  k  =  1. 

120.  A  point  moves  so  that  the  sum  of  the  squares  of  its  dis- 
tances from  the  sides  of  an  equilateral  triangle  is  constant.  Show 
that  the  locus  is  a  circle  and  find  its  center. 

121.  A  f>oint  moves  so  that  the  square  of  its  distance  from  the 
base  of  an  isosceles  triangle  is  equal  to  the  product  of  its  distances 
from  the  other  two  sides.  Show  that  the  locus  is  a  circle  and  an 
hyperbola  which  pass  through  the  vertices  of  the  two  base  angles. 

122.  A  point  moves  so  that  the  sum  of  the  squares  of  its  dis- 
tances from  the  four  sides  of  a  square  is  constant.    Find  its  locus. 

123.  A  point  moves  so  that  the  sum  of  the  squares  of  its  dis- 
tances from  any  number  of  fixed  points  is  constant.    Find  its  locus. 

124.  Find  the  locus  of  a  point  the  square  of  the  distance  of 
which  from  a  fixed  point  is  proportional  to  its  distance  from  a 
fixed  straight  line. 

125.  Find  the  locus  of  a  point  such  that  the  lengths  of  the  tan- 
gents from  it  to  two  concentric  circles  are  inversely  as  the  radii  of 
the  circles. 


PROBLEMS  101 

126.  A  point  moves  so  that  the  length  of  the  tangent  from  it  to 
a  fixed  circle  is  equal  to  its  distance  from  a  fixed  point.  Find  its 
locus. 

127.  Find  the  locus  of  a  point  the  tangents  from  which  to  two 
fixed  circles  are  of  equal  length. 

128.  Straight  lines  are  drawn  through  the  points  (—  a,  0)  and 
(a,  0)  so  that  the  difference  of  the  angles  they  make  with  the  axis 

of  x  is  tan-1--    Find  the  locus  of  their  point  of  intersection. 

129.  The  slope  of  a  straight  line  passing  through  (a,  0)  is  twice 
the  slope  of  a  straight  line  passing  through  (—  a,  0).  Find  the  locus 
of  the  point  of  intersection  of  these  lines. 

130.  A  point  moves  so  that  the  product  of  the  slopes  of  the 
straight  lines  joining  it  to  A  (—  a,  0)  and  /;  (a,  0)  is  constant. 
Prove  that  the  locus  is  an  ellipse  or  an  hyperbola. 

131.  If,  in  the  triangle  ABC,  tan  .1  tan  \  11  =  1'  and  Ml  is  fixed, 
show  that  the  locus  of  C  is  a  parabola  with  its  vertex  at  A  and  its 
focus  at  B. 

132.  Given  the  base  2b  of  a  triangle  and  the  sum  s  of  the  tan- 
gents of  the  angles  at  the  base.    Find  the  locus  of  the  vertex. 

133.  Find  the  locus  of  the  center  of  a  circle  which  is  tangent  to 
a  fixed  circle  and  a  fixed  straight  line. 

134.  Prove  that  the  locus  of  the  center  of  a  circle  which  passes 
through  a  fixed  point  and  is  tangent  to  a  fixed  straight  line  is  a 
parabola. 

135.  A  point  moves  so  that  its  shortest  distance  from  a  fixed 
circle  is  equal  to  its  distance  from  a  fixed  diameter  of  that  circle. 
Find  its  locus. 

136.  If  a  straight  line  is  drawn  from  the  origin  to  any  point  Q  of 
the  line  y  =  a,  and  if  a  point  P  is  taken  on  this  line  such  that  its 
ordinate  is  equal  to  the  abscissa  of  Q,  find  the  locus  of  P. 

137.  AOB  and  COD  are  two  straight  lines  which  bisect  each  other  at 
right  angles.   Find  the  locus  of  a  point  P  such  that  PA  ■  PB  =  PC  •  PD. 

138.  AB  and  CD  are  perpendicular  diameters  of  a  circle  and  M  is 
any  point  on  the  circle.  Through  M,  AM  and  BM  are  drawn.  AM 
intersects  CD  in  N,  and  from  7Y  a  straight  line  is  drawn  parallel  to 
AB,  meeting  BM  in  P.    Find  the  locus  of  P. 


102  CERTAIN"  CURVES 

139.  Given  a  fixed  straight  line  All  and  a  fixed  point  Q.  From 
any  point  R  in  AB  a  perpendicular  to  AB  is  drawn,  equal  in  length 
to  RQ.    Find  the  locus  of  the  end  of  this  perpendicular. 

140.  0  is  a  fixed  point  and  AB  is  a  fixed  straight  line.  A  straight 
line  is  drawn  from  0,  meeting  AB  at  Q,  and  in  OQ  a  point  P  is  taken 
so  that  OP  -OQ=  k2.    Find  the  locus  of  P. 

141.  Let  OA  be  the  diameter  of  a  fixed  circle.  From  B,  any  point 
on  the  circle,  draw  a  straight  line  perpendicular  to  OA,  meeting  it 
in  D.  Prolong  the  line  DB  to  P,  so  that  OD:DB=  OA :  DP.  Find 
the  locus  of  P. 

142.  A  perpendicular  is  drawn  from  the  focus  of  an  hyperbola  to 
an  asymptote.  Show  that  its  foot  is  at  distances  a  and  b  from  the 
center  and  the  focus  respectively. 

143.  Two  straight  lines  are  drawn  through  the  vertex  of  a  parab- 
ola at  right  angles  to  each  other  and  meeting  the  curve  at  P  and  Q. 
Show  that  the  line  PQ  cuts  the  axis  of  the  parabola  in  a  fixed  point. 

144.  In  the  parabola  if  =  A,px  an  equilateral  triangle  is  so 
inscribed  that  one  vertex  is  at  the  origin.  What  is  the  length 
of  one  of  its  sides  ? 

145.  Prove  that  in  the  ellipse  half  of  the  minor  axis  is  a  mean 
proportional  between  A  F  and  FA '. 

146.  Show  that  in  an  equilateral  hyperbola  the  distance  of  a  point 
from  the  center  is  a  mean  proportional  between  the  focal  distances  of 
the  point. 

147.  If  from  any  point  P  of  an  hyperbola  PK  is  drawn  parallel 
to  the  transverse  axis,  cutting  the  asymptotes  in  Q  and  R,  prove 
PQ  •  PR  =  «2.  If  PK  is  drawn  parallel  to  the  conjugate  axis,  prove 
PQ  .  PR  =  _  62. 

148.  Prove  that  the  product  of  the  distances  of  any  point  of  the 
hyperbola  from  the  asymptotes  is  constant. 

149.  Prove  that  in  the  hyperbola  the  squares  of  the  ordinates  of 
any  two  points  are  to  each  other  as  the  products  of  the  segments  of 
the  transverse  axis  made  by  the  feet  of  these  ordinates. 

150.  Straight  lines  are  drawn  through  a  point  of  an  ellipse  from 
the  two  ends  of  the  minor  axis.  Show  that  the  product  of  their 
intercepts  on  OX  is  constant. 


PROBLEMS 


103 


151.  P1  is  any  point  of  the  parabola  y2  =  4^.r,  and  PXQ,  which 
is  perpendicular  to  OPv  intersects  the  axis  of  the  parabola  in  Q. 
Prove  that  the  projection  of  PXQ  on  the  axis  of  the  parabola  is 
always  4^>. 

152.  Show  that  the  focal  distance  of  any  point  on  the  hyperbola, 
is  equal  to  the  length  of  the  straight  line  drawn  through  the  point 
parallel  to  an  asymptote  to  meet  the  corresponding  directrix. 

153.  Show  that  the  following  points  lie  approximately  on  a  straight 
line,  and  find  its  equation : 


X 

4 

9 

13 

20           22 

25 

30 

?/ 

2.1 

4.6 

7 

12          12.0 

14.5 

18.2 

154.  For  a  galvanometer  the  deflection  />,  measured  in  millimeters 
on  a  proper  scale,  and  the  current  /.  measured  in  microamperes,  are 
determined  in  a  series  of  readings  as  follows  : 


J) 

29.1 

48.2 

72.7 

92.0 

IIS. II 

140.0 

i  <;.-..<  i 

199.0 

I 

0.0493 

0.0821 

0.123 

0.164 

0.197 

0.234 

0.274 

0.328 

Find  an  empirical  law  connecting  1>  and  /. 

155.  For  a  copper-nickel  thermocouple  the  relation  between  the 
temperature  t  in  degrees  and  the  thermoelectric  power  in  microvolts 
is  given  by  the  following  table  : 


t 

0 

50  '       100     !      150 

200 

p 

24 

2.".           26         26.9 

27.6 

Find  an  empirical  law  connecting  t  and  p 


156.  The  safe  loads  in  thousands  of  pounds  for  beams  of  the  same 
cross-section  but  of  various  lengths  in  feet  are  found  as  follows  : 


Length 

10 

11 

12 

13               14 

15 

Load 

123.6 

121.5 

111.8 

107.2 

101.3 

90.4 

Find  the  empirical  equation  connecting  the  data. 


104 


CERTAIN  CURVES 


157.  The  relation  between  the  pressure  p  and  the  volume  v  of  a 
gas  is  found  experimentally  as  follows : 


Pressure 

20 

23.5 

31 

42 

59 

78 

Volume 

0.619 

0.540 

0.442 

0.358 

0.277 

0.219 

Find  an  empirical  equation  connecting />  and  v  in  the  form  pvn  =  c. 

158.  The  deflection  a  of  a  loaded  beam  with  a  constant  load  is 
found  for  various  lengths  I  as  follows : 


I 

1000 

900 

800 

700 

600 

a 

7.14 

5.22 

3.64 

2.42 

1.50 

Find  an  empirical  equation  connecting  a  and  I  in  the  form  a  =  JcF. 

159.  The  relation  between  the  length  I  (in  mm.)  and  the  time  t 
(in  seconds)  of  a  swinging  pendulum  is  found  as  follows  : 


I 

63.4 

80.5 

90.4 

101.3 

107.3 

140.6 

t 

0.806 

0.892 

0.960 

1.010 

1.038 

1.198 

Find  an  empirical  equation  connecting  I  and  t  in  the  form  t  =  Id'1. 
160.  For  a  dynamometer  the  relation  between  the  deflection  6, 


as  follows  : 


J7T 

400 L 


6 

40 

86 

120 

160 

201 

240 

280 

320 

362 

I 

0.147 

0.215 

0.252 

0.293 

0.329 

0.360 

0.390 

0.417 

0.442 

Find  an  empirical  equation  connecting  /  and  $  in  the  form  I  =  k$". 

161.  In  a  chemical  experiment  the  relation  between  the  concen- 
tration y  of  undissociated  hydrochloric  acid  is  connected  with  the 
concentration  x  of  hydrogen  ions  as  shown  in  the  table : 


X 

1.68 

1.22 

0.784 

0.426 

0.092 

0.047 

0.0096 

0.0049 

0.00098 

y 

1.32 

0.676     0.216 

0.074    0.0085 

0.00315 

0.00036 

0.00014 

0.000018 

Find  an  empirical  law  connecting  the  two  quantities  in  the  form  y = kxn 


PROBLEMS 


105 


162.  Show  that  the  values  of  x  and  y  as  given  in  the  following 
table  are  connected  by  a  relation  of  the  form  y  =  cax,  and  find  c  and  a. 


X 

8 

10 

12      '      14 

16 

18 

20 

y 

3.2 

4.6 

7.3 

9.8 

15.2 

24.6 

36.4 

163.  In  a  certain  chemical  reaction  the  concentration  c  of  sodium 
acetate  produced  at  the  end  of  the  stated  number  of  minutes  t  is 

as   follows  : 


t 

1 

2 

3                      4 

c 

0.00837 

0.0070 

0.00586     J     0.00402         0.00410 

Assuming  that  the  law  is  of  the  form  <■  —  ab*}  find  the  equation 
connecting  the  concentration  with  the  time. 

164.  The  molal  heat  capacity  at  constant  temperature  is  for  water 
vapor  at  various  temperatures  as  follows: 


Temp. 

10 

100 

500 

700 

1000 

Cap. 

8.8 

8.0 

8.4 

8.6 

9.1 

Determine  the  law  in  the  form  C  =  a  +  bt  +  rf~- 

165.  Assuming  Boyle's'law,  pv  =  c,  determine  <■  graphically  from 
the  following  pairs  of  observed  values: 


V 

39.02 

42.17 

45.80 

18.52 

51.89 

60.47 

65.97 

V 

40.37 

38.32 

85.32 

33.29 

31.22 

26.86 

24.53 

166.  The  distance  p  of  an  object  from  a  lens  and  the  distance  p1 
of  its  image  are  found  by  experiment  as  follows  : 


p 

320 

240 

180 

140 

120 

100 

80 

60 

p' 

21.35 

21.80 

22.50 

23.20 

23.80        24.60 

26.20 

29.00 

here  f  is  the  focal  length  of  the  lens, 


Assuming  the  law  -  -\ :  =  — 

P       P       ./ 
compute  /  graphically  by  plotting  the  reciprocals  of  p  and  p' 


CHAPTER  VII 


PARAMETRIC  REPRESENTATION 


52.  Definition.    Consider  the  two  equations 


(i) 


where  ^(0  and  f2(f)  are  two  functions  of  an  independent 
variable  t.  If  we  assign  to  t  any  value  in  (1),  we  determine 
x  and  y  and  may  plot  a  point  with  these  coordinates.  In 
this  way  a  value  of  t  determines  a  point  in  the  plane.  So 
other  values  of  t  determine  other  points,  which  together  deter- 
mine a  curve. 

The  two  equations  (1)  then  represent  the  curve.  The  vari- 
able t  is  called  a  parameter,  and  the  equations  (1)  are  called 
the  parametric  representation  of  the  curve.  It  is  sometimes 
easy  to  eliminate  t  from  the  equations  (1)  and  obtain  thus 
a  Cartesian  equation 
of  the  curve,  but  this  y 
elimination  is  not  essen- 
tial and  is  not  always 
desirable. 


Ex.  1.    x  =  t'2,  y 

Giving  t  in 
the  values  —  3,  - 
1,  2,  3,  we  find 
sponding      points 


=  t. 

succession 
2,  -1,  0, 
the  corre- 
(9,      -3), 


Fig.  88 


(4,  -2),  (1,  -1),  (0,  0), 
(1,  1),  (4,  2),  (9,  3).  These 
points,  if  plotted,  may  be 
connected  by  the  curve  of 
fig.  88,  and  as  many  inter- 
mediate points  as  desired  may  be  found.  In  this  case  we  may  easily 
eliminate  t  from  the  equations  and  obtain  x  =  y2.    The  curve  is  a  parabola. 

106 


PARAMETRIC  REPRESENTATION 


107 


Ex.  2.  x  =  t3  +  2  fi,  y 


|,   -1,   -i,   0,    i,   1,  we 
•),   (1,  0),   (|,    |),   (0,  0), 


Giving  t  in  succession  the  values  —  2,  - 
find  as  corresponding  points  (0,  —  (J),  {%,  — 
(J,  -  t)>  (3,  0). 

These  points  give  the  curve  shown  in 
fig.  89.  If  more  details  as  to  the  shape  of 
the  loop  are  wanted,  more  values  of  t  must 
be  assumed  intermediate  to  those  we  have 
used.  Elimination  of  /  in  this  example  is 
possible  hut  hardly  desirable. 


Ex.  3.  ./■  =  <i  ens"/,   ij  =  */  sin8tf. 

If  values  of  t  are  assumed  at  convenient 
intervals  between  t  =  0°  and  t  =  360°,  the 
curve  may  he  found  to  be  as  in  fig.  !»(>. 
The  elimination  of  t  gives  the  equation 
£j  +  yt  =  nt.  The  curve  is  called  the  four- 
etaped  hypocycloid  (§  58). 

As  the  examples  show,  the  param- 
eter t  is  in  general  simply  an  inde- 
pendent variable   to  which  values  are 

assigned  at  pleasure.  In  problems  of  mechanics,  however,  the  pa- 
rameter frequently  represents  time.  In  this  ease  the  curve  of  equa- 
tions (1)  represents  the  path 
of  a  moving  point,  the  position 
of  the  point  at  any  instant 
being  given  by  the  equations. 
Any  of  the  above  examples 
may  be  interpreted  in  this 
way.  Other  illustrations  will 
be  found  in  the  examples  of 
§§53  and  54. 

In  some  cases,  also,  it  is 
possible  to  give  a  geometric 
interpretation  to  the  param- 
eter t.  This  is  illustrated 
by  the  curves  which  follow,  where  in  each  case  the  parameter  is 
a  certain  angle. 


108 


I .» A 1 ;  A  M  ETRIC   REPRESENTATION 


53.  The  circle.  Let  P(x,  y)  (fig.  91)  be  any  point  on  a  circle 
with  its  center  at  the  origin  0  and  its  radius  equal  to  a.  Let  </> 
be  the  angle  made  by  OP  and  OX.  Then,  from  the  definition  of 
the  sine  and  cosine, 

x  =  a  cos  cj>, 

y  =  a  sin  c/>, 

are  the  parametric  equations  of  the  circle 
with  <f>  as  the  arbitrary  parameter. 

Ex.  A  particle  moves  in  a  circle  at  a  con- 
stant rate  k.  Then,  if  s  represents  the  arc 
traversed  in  the  time  /, 

s  —  kt     and     <h  =  -  =  — 


Therefore  the  equations  of  the  circle  are 
kt 


Fig.  91 


54.  The  ellipse.    In  a  circle  with  radius  a  let  the  abscissa  of 
every  point  Q  (fig.  92)  be  left  unchanged  and  its  ordinate  be 
altered  in  a  fixed  ratio  b :  a,  where  b  is  any  length  whatever. 
The  point  Q  then  takes  such  a  position 
as  P,  where  in  the  figure  b  <  a.    The 
parametric  equations  of  the  locus  of  P 
are  therefore,  from  §  53, 

x  =  a  cos  </>, 

y  =  b  sin  </>. 

The  elimination  of  cf>  from  these  equa- 


tions gives  I- 


+[i 


1,  showing  that 


Fig.  92 


the  locus  of  P  is  an  ellipse. 

4>  is  called  the  eccentric  angle  of  a  point  on  the  ellipse,  and 
the  circle  x2+y2=  a2  is  called  the  auxiliary  circle. 

Ex.    A  particle  Q  moves  at  a  constant  rate  along  the  auxiliary  circle  of 
an  ellipse ;  required  the  motion  of  its  accompanying  point  P. 
As  in  §  53,  <f>  =  —■    Hence  the  equations  of  the  path  are 


kt 


y 


bsm 


It 


THE  CYCLOID 


109 


55.  The  cycloid.  If  a  circle  rolls  upon  a  straight  line,  each 
point  of  the  circumference  describes  a  curve  called  a  cycloid. 

Let  a  circle  of  radius  a  roll  upon  the  axis  of  x,  and  let  C 
(fig.  93)  be  its  center  at  any  time  of  its  motion,  N  its  point  of 
contact  with  OX,  and  P  the  point  on  its  circumference  which 


Fig.  93 


describes  the  cycloid.    Take  as  the  origin  of  coordinates,  0,  the 
point  found  by  rolling  the  circle  to  the  left  until  P  meets  OX. 

Then  ON=  arc  I'X. 

Draw  MP  and  CX,  each  perpendicular  to  OX,  PR  parallel  to 
OX,  and  connect  C  and  P.    Let 

angle  X('P=  <f>. 
Then  x  =  O M  =  OX  -  MN 

=  arc  XI'- PR 
=  a  <f>  —  a  sin  (f>. 
ij  =  MI>  =  XC-RC 
=  a  —  a  cos  (f>. 

Hence  the  parametric  representation  of  the  cycloid  is 

x=  a($  —  sin  </>), 

y  =  a  (1  —  cos  <£). 
By  eliminating  <£,  the  equation  of  the  cycloid  may  be  written 

_    „-ia— y 


_V2  ay  -  y\ 

but  this  is  less  convenient  than  the  parametric  representation. 

At  each  point  where  the  cycloid  meets  OX  a  sharp  vertex 
called  a  cusp  is  formed.  The  distance  between  two  consecutive 
cusps  is  evidently  2  ira. 


110 


PARAMETRIC   I  IK  PRESENTATION 


56.  The  trochoid.  When  a  circle  rolls  upon  a  straight  line, 
any  point  upon  a  radius,  or  upon  a  radius  produced,  describes 
a  curve  called  a  trochoid. 

Let  the  circle  roll  upon  the  axis  of  x,  and  let  C  (figs.  9-4  and 
95)  be  its  center  at  any  time,  N  its  point  of  contact  with  the 


axis  of  x,  P(x,  y)  the  point  which  describes  the  trochoid,  and 
K  the  point  in  which  the  line  CP  meets  the  circle.  Take  as 
the  origin  0  the  point  found  by  rolling  the  circle  toward  the 
left  until  A'  is  on  the  axis  of  x.     Then 

ON  =  arc  NX. 


Draw  PM  and  CN  perpendicular  to  OX,  and  through  P  a  line 
parallel  to  OX,  meeting  CN,  or  CN  produced,  in  R.  Let  the 
radius  of  the  circle  be  a,  CP  be  h,  and  angle  NCP  be  <f).    Then 

x=OM=ON-MN 

=  arc  NK—  PR 

=  d(f)  —  h  sin  (p. 
y  =  MP  =  NC  -  RC 


a 


h  cos 


THE  EPICYCLOID 


111 


57.  The  epicycloid.  When  a  circle  rolls  upon  the  outside  of  a 
fixed  circle,  each  point  of  the  circumference  of  the  rolling  circle 
describes  a  curve  called  an  epicycloid. 

Let  0  (fig.  96)  be  the  center  of  the  fixed  circle,  0  the  center 
of  the  rolling  circle,  K  its  point  of  contact  with  the  fixed  circle, 
and  P(x,  y)  the  point 
which  describes  the 
epicycloid.  Determine 
the  point  K  by  rolling 
the  circle  C  until  P 
meets  the  circumference 
of  0.    Then 

arc  KN=  arc  NP. 

Take  0  as  the  origin 
of  coordinates  and  OK 
as  the  axis  of  x.  Draw 
PM  and  CL  perpendic- 
ular to  OX,  PS  parallel 
to  OX,  meeting  CL  in 
E,  and  connect  0  and  C. 
Let    the   radius   of    the 

rolling  circle  be  a.  that   of  tin-  fixed  circle  ?>,  ami  denote  the 
angle  OCP  by  0,  the  angle  KOC  by  cf>.    Thru 

arc  KX=  /-</>,  an-  NP  =  ad ; 

whence  /«/>  =  ad. 

We  now  have    x  =  031=  OL+L  .V 

=  OC  cos  KOC-CP  cos  SPC 
=  (a  +  li)  cos  <f>  —  a  cos  (<£  +  0) 
a  +  b 
a 
y  =  MP  =  LC-EC 
=  0C  sin  KOC  -  OP  sin  SPC 
=  (a  +  b)  sin  <£  —  a  sin  ($  +  0) 

=  (a  +  b)  sin  $  —  a  sin <£>. 


=  (a  +  A)  cos  ^-aci  >s 


</>• 


112 


PARAMETRIC  R  EP RES  EKT ATION 


The  curve  consists  of  a  number  of  congruent  arches,  the  first  of 
which  corresponds  to  values  of  6  between  0  and  2  7r,  that  is,  to 


Similarly,  the  Mi  arch  corre- 


2(k 


1)  air        ,  2  kair      TT 

and  — : — •    Hence 


values  of  <£  between  0  and  — — 

spon ds  to  values  of  <j>  between 

b  o 

the  curve  is  a  closed  curve  when,  and  only  when,  for  some  value 

of  k,  — —  is  a  multiple  of  2  it.    \ia  and  b  are  incommensurable, 
b 

this  is  impossible,  but  if  -  =  *1 ,  where  —  is  a  rational  fraction  in 

its  lowest  terms,  the  smallest  value  of  k  —  q.   The  curve  then  con- 
sists of  q  arches  and  winds  p  times  around  the  fixed  circle. 

58.  The  hypocycloid.  When  a  circle  rolls  upon  the  inside  of 
a  fixed  circle,  each  point  of  the  rolling  circle  describes  a  curve 
called  the  hypocycloid.  If  the  axes  and  the  notation  are  as  in 
the  previous  article,  the  equa- 
tions of  the  hypocycloid  are 

x  =  (b  —  a.)  cos  (f>  +  a  cos  — 


y  =  (b  —  a)  sin  </>  —  a  sin ■  <fi 


—  a 
a 
b-a 


The  proof  is  left  to  the  student. 
The  curve  is  shown  in  fig.  97. 

In  the  special  case  in  which 
the  radius  of  the  rolling  circle 
is  one  fourth  that  of  the  fixed 
circle,  we  have  b  =  4  a.    Then  FlG-  97 

x  =  a  (3. cos  (f>  +  cos  3  $)  =  4  a  cos3<£  =  b  cos3<£, 
y  —  a  (3  sin  c/>  —  sin  3  (/>)  =  4  a  sin3(£  =  b  sin3<£. 

This  is  the  four-cusped  hypocycloid  of  Ex.  3,  §  52. 

59.  The  involute  of  the  circle.  If  a  string,  kept  taut,  is 
unwound  from  the  circumference  of  a  circle,  its  end  describes 
a  curve  called  the  involute  of  the  circle.  Let  0  (fig.  98)  be 
the  center  of  the  circle,  a  its  radius,  and  A  the  point  at  which 


PROBLEMS 


113 


the  end  of  the  string  is  on  the  circle.  Take  0  as  the  origin  of 
coordinates  and  OA  as  the  axis  of  x.  Let  P(x,  y)  be  a  point 
on  the  involute,  PK  the  line  drawn  from  P  tangent  to  the 
circle  at  K,  and  <£  the  angle 
XOK.  Then  PK  represents  a 
portion  of  the  unwinding  string, 
and  hence 

KP  —  arc  AK=atf>. 

Now   it   is   clear   that   for   all 
positions    of    the    point   K,    OK 

makes  an  angle  <f>  —  -^  with  OY. 

Hence  the  projection  of  OK  on 
OX  is  always  OK  cos  <j>  =  a  cos  <j>, 
and    its    projection    on    OY   is  F"  •  '-,s 

(7j-\  jj- 

(f)  —  —  \  =  a  sin  <f>.  Also  KP  always  makes  an  angle  (f>  —  — 

with  OX  and  an  angle  ir  —  <f>  with  (>  Y.    Hence  the  projection  of 

KP   on    OX   is    KP  cos  I  <f>  —  — )  =  atf>  sin  <£,    and    its    projection 

on  C>  F  is  jKP  cos  (7r  —  <£)  =  —  «</>  cos  (/>.   The  projection  of  OP  on 
OX  is  #,  and  on  01' is  y.    Hence,  by  the  law  of  projections,  §  2, 

x  =  a  cos  <$>  +  d(f>  sin  (f>, 
y  =  a  sin  (f>—  acf)  cos  <f). 


PROBLEMS 

Plot  the  graphs  of  the  following  parametric  equations : 

6  6t 


1.  x  =  t2,  y  —  t  -f-  1. 


4  4 

2.  x  =  —  >    w  =  -• 
*2     lX        £ 


3.    £C  = 


±  Vl  +  9 11 


.'/ 


±  Vi  +  9  e 


4.  x  =  f,   y 


a2  +  *2 


5 .  x  =  - o  >    1 1  = 

l  +  t-  J 

6.  x  =  2  a  shr<f>,  y 

7 .  cr  =  e*  sin  £,  ?/  = 


t(l+l*) 
2  a  sin3<£ 
cos  <£ 
e*  cos  £. 


x  =  a<f>  +  a  sin  cf>,   ij  =  r<  ■ 

—  a-  cos  e£. 

3  a                  a         Q 

iC  —  —£-  cos  4>  —  -x  cos  3  <£, 

3  a    . 
y  =  —  sin  <£ 

a'  =  2  a  cos  <j>  —  a  cos  2  <£, 

y  =  2  a  sin  <£ 

114  PARAMETRIC  REPRESENTATION 


-  sin  3  <£. 

10.  a*  =  2  a  cos  <£  —  a  cos  2  <f>,  y  =  2  a  sin  <f>  —  a  sin  2  <£. 

11.  A  projectile  moves  so  that  the  coordinates  of  its  position 
at  any  time  t  are  given  by  the  equations  x  =  Q0t,*y=  80£  — 16Z2. 
Plot  its  path. 

12.  Find  the  parametric  equations  of  the  parabola  y2  =  4tpx  when 
the  parameter  is  the  slope  of  a  straight  line  through  the  vertex. 

x2       ?/2 

13.  Find  the  parametric  equations  of  the  ellipse  —  +  —  =1  when 

the  parameter  is  the  slope  of  a  straight  line  through  the  center. 

14.  Find  the  parametric  equations  of  the  cissoid  when  the  param- 
eter is  the  slope  of  a  straight  line  through  the  origin,  the  axes  of 
coordinates  being  as  in  fig.  81. 

15.  Find  the  parametric  equations  of  the  cissoid  when  the  param- 
eter is  the  angle  A  OP  (fig.  81). 

16.  Find  the  parametric  equations  of  the  strophoid  when  the 
parameter  is  the  angle  MAP  (fig.  82). 

17.  When  a  circle  rolls  upon  the  outside  of  a  fixed  circle,  a  point 
on  the  radius  of  the  rolling  circle  at  a  distance  h  from  its  center 
describes  a  curve  called  an  epitrochoid.    Find  its  equations. 

18.  When  a  circle  rolls  upon  the  inside  of  a  fixed  circle,  a  point 
on  the  radius  of  the  rolling  circle  at  a  distance  h  from  its  center 
describes  a  curve  called  an  hypotrochoid.    Find  its  equations. 

19.  If  a  circle  rolls  on  the  inside  of  a  fixed  circle  of  twice  its 
radius,  what  is  the  form  of  the  curve  generated  by  a  point  of  the 
circumference  of  the  rolling  circle  ? 

20.  AB  is  a  given  straight  line  perpendicular  to  OX  at  the  point 
C,  where  OC  —  a.  Through  0  any  straight  line  is  drawn,  meeting  AB 
at  D.  On  OX  a  point  M  is  taken,  to  the  left  of  C,  so  that  CM  =  CD. 
Finally,  through  M  a  straight  line  is  drawn  perpendicular  to  OX, 
intersecting  OD  at  P.  Find  the  parametric  equations  of  the  locus  of 
P,  using  the  angle  XOD  as  the  parameter.  Find  also  the  Cartesian 
equation,  name  the  curve,  and  sketch  the  graph. 


PROBLEMS  115 

21.  A  fixed  circle  of  radius  a  with  its  center  at  0  intersects  OX 
at  A.  The  straight  line  BC  is  tangent  to  the  circle  at  .4.  Through  0 
any  straight  line  is  drawn,  intersecting  the  circle  at  D  and  intersecting 
BC  at  E.  Through  D  a  straight  line  is  drawn  parallel  to  OY,  and 
through  E  a  straight  line  is  drawn  parallel  to  OX.  These  lines  inter- 
sect at  P.  Find  the  parametric  equations  of  the  locus  of  P  in  terms 
of  the  angle  XOD  as  parameter.  Find  also  the  Cartesian  equation 
and  sketch  the  curve. 

22.  A  circle  of  radius  a  has  its  center  at  0,  the  origin  of  coordinates. 
The  tangent  to  the  circle  at  any  point  A  meets  OX  at  .1/.  Through  M 
a  straight  line  is  drawn  parallel  to  <>Y,  and  through  .1  a  straight  line 
is  drawn  parallel  to  OX.  These  lines  intersect  at  P.  Find  the  par- 
ametric equations  of  the  locus  of  P,  using  the  angle  MOA  as  the 
parameter.    Find  also  the  Cartesian  equation  and  sketch  the  curve. 

23.  A  circle  of  radius  a  has  its  center  at  the  origin  of  coordinates 
0.  Through  O  any  straight  line  is  drawn,  intersecting  the  circle  at 
A.  The  tangent  to  the  circle  at  .1  intersects  OFat  /;.  Through  B  a 
straight  line  is  drawn  parallel  to  0X}  meeting  OA  produced  at  /'. 
Find  the  parametric  equations  of  the  locus  of  P  in  terms  of  the 
angle  XOA  as  parameter.    Find  also  the  Cartesian  equation. 

24.  Let  OA  be  the  diameter  of  a  fixed  circle  and  LK  the  tangent 
at  A.  From  O  draw  any  straight  line  intersecting  the  circle  at  B 
and  LK  at  C,  and  let  /'  be  the  middle  point  of  BC.  Find  the  para- 
metric equations  of  the  locus  of  /',  using  the  angle  AOP  as  the 
parameter,  OA  as  the  axis  of  y,  and  0  as  the  origin.  Find  also  the 
Cartesian  equation. 

25.  A  circle  of  radius  a  has  its  center  at  the  origin  of  coordinates 
0,  and  the  straight  line  -I  B  is  tangent  to  the  circle  at  A(a,  0).  From 
0  any  straight  line  is  drawn,  meeting  All  at  /•;  and  the  circle  at  D. 
On  OE,  OP  is  taken  equal  to  DE.  Find  the  parametric  equations  of 
the  locus  of  P  in  terms  of  the  angle  AOP  as  parameter. 

26.  The  straight  line  All  is  perpendicular  to  OX  at  A  (a,  0).  From 
0  a  straight  line  is  drawn  to  any  point  C  of  All.  The  straight  line 
drawn  from  C  perpendicular  to  OC  meets  OX  at  M.  The  perpendic- 
ular to  OX  at  .1/  meets'  OC  produced  at  P.  Find  the  parametric 
equations  of  the  locus  of  P  in  terms  of  the  angle  XOC  as  parameter. 
Find  also  the  Cartesian  equation. 


116  PARAMETRIC  REPRESENTATION 

27.  OBCD  is  a  rectangle  with  OB  =  a  and  BC  =  c.  Any  line  is 
drawn  through  C,  meeting  OB  in  E,  and  the  triangle  EPO  is  con- 
structed so  that  the  angles  CEP  and  EPO  are  right  angles.  Find 
the  parametric  equations  of  the  locus  of  P,  using  the  angle  DOP  as 
I  he  parameter,  OB  as  the  axis  of  x,  and  0  as  the  origin.  Find  also  the 
Cartesian  equation  of  the  locus. 

28.  A  fixed  circle  has  as  diameter  the  straight  line  joining  the 
origin  and  the  point  A  (0,  2a).  Any  point  B  of  the  circle  is  connected 
with  A  and  0,  and  BM  is  drawn  perpendicular  to  OX,  meeting  OX 
at  M.  On  MB,  MP  is  laid  off  equal  to  BA.  Find  the  parametric 
equations  of  the  locus  of  P  in  terms  of  the  angle  XOB  as  parameter. 
Find  also  the  Cartesian  equation. 

29.  Let  AB  be  a  given  straight  line,  0  a  given  point  a  units  from 
AB,  and  k  a  given  constant.  On  any  straight  line  through  0,  meet- 
ing AB  in  M,  take-P  so  that  OM  •  MP=  k2.  Find  the  parametric 
equations  of  the  locus  of  P,  using  0  as  the  origin,  the  perpendicular 
from  0  to  AB  as  the  axis  of  x,  and  the  angle  between  OX  and  OP 
as  the  parameter.   Also  find  the  Cartesian  equation. 

30.  ABC  is  a  given  right  triangle  of  which  the  sides  AB  and  BC 
about  the  right  angle  at  B  are  always  equal  to  a  and  b  respectively. 
The  triangle  moves  in  the  plane  XOY  so  that  A  is  always  on  OF  and 
B  is  always  on  OX.  P  is  the  middle  point  of  the  hypotenuse  AC. 
Find  the  parametric  equations  of  the  locus  of  P,  using  the  angle 
XBC  as  the  parameter. 

31.  Let  0  be  the  center  of  a  circle  with  radius  a,  A  a  fixed  point 
on  the  circle,  and  B  a  moving  point  on  the  circle.  If  the  tangent  at 
B  meets  the  tangent  at  A  in  C,  and  P  is  the  middle  point  of  BC, 
find  the  equations  of  the  locus  of  P  in  parametric  form,  using  the 
angle  A  OB  as  the  arbitrary  parameter,  OA  as  the  axis  of  x,  and  0 
as  the  origin. 

32.  A  fixed  circle  has  as  diameter  the  straight  line  joining  the 
origin  of  coordinates  and  the  point  A  (2  a,  0),  and  LK  is  tangent  to 
the  circle  at  A.  From  O  any  straight  line  is  drawn,  meeting  the  circle 
at  D  and  the  tangent  LK  at  E.  On  OE  a  point  P  is  so  taken  that 
PD  =  DE  in  both  length  and  direction.  Find  the  parametric  equa- 
tions of  the  locus  of  P  in  terms  of  the  angle  AOE.&s  parameter. 
Find  also  the  Cartesian  equation. 


PROBLEMS  117 

33.  A  and  B  are  two  points  on  the  axis  of  y  at  distances  —  a  and 
-f-  a  respectively  from  the  origin.  AH  is  any  straight  line  through  A, 
meeting  the  axis  of  x  at  H.  BK  is  the  perpendicular  from  B  on  All, 
meeting  it  at  K.  Through  K  a  straight  line  is  drawn  parallel  to 
the  axis  of  x,  and  through  H  a  straight  line  is  drawn  parallel  to 
the  axis  of  y.  These  lines  meet  in  P.  Find  the  parametric  equa- 
tions of  the  locus  of  P,  using  the  angle  BAK  as  the  parameter. 
Also  find  the  Cartesian  equation. 

34.  Q  is  the  point  on  the  auxiliary  circle  of  the  ellipse 

corresponding  to  the  point  P  of  the  ellipse.  The  straight  line 
through  P  parallel  to  OQ  meets  OX  at  L  and  OY  at  M.  Prove 
PL  =  b,  and  PM=  a. 

35.  If  a  projectile  starts  with  an  initial  velocity  v  in  an  initial 
direction  which  makes  an  angle  a  with  the  axis  of  x,  taken  horizontal, 
its  position  at  any  time  t  is  given  by  the  parametric  equations 

x  =  vt  cos  a,         y  =  vt  sin  a  —  \  yP. 

Pind  the  Cartesian  equation  of  the  path  of  the  projectile  and  its 
nature  and  position. 

36.  From  the  equations  of  problem  35  determine  when  and  where 
the  projectile  strikes  a  point  on  the  axis  of  x. 

37.  From  the  equations  of  problem  35  determine  when,  and  for 
what  value  of  x,  the  projectile  passes  through  a  point  which  is  at 
a  distance  h  below  the  horizontal. 

38.  From  the  equations  of  problem  35,  what  elevation  must  be 
given  to  a  gun  that  the  projectile  may  pass  through  a  poinl  b  units 
distant  from  the  muzzle  of  the  gun  and  lying  in  the  horizontal  line 
passing  through  the  muzzle  ? 

39.  From  the  equations  of  problem  35,  what  elevation  must  be 
given  to  a  gun  to  obtain  a  maximum  range  on  a  horizontal  line 
passing  through  the  muzzle  ? 

40.  A  gun  stands  on  a  cliff  //  units  above  the  water.  From  the 
equations  of  problem  35,  what  elevation  must  be  given  to  the  gun 
that  the  projectile  may  strike  a  point  in  the  water  b  units  from  the 
base  of  the  cliff  ? 

AC 


CHAPTER  VIII 
POLAR  COORDINATES 

60.  Coordinate  system.  So  far  we  have  determined  the  posi- 
tion of  a  point  in  the  plane  by  two  distances,  x  and  y.  We 
may,  however,  use  a  distance  and  a  direction,  as  follows: 

Let  0  (fig.  99),  called  the  origin,  or  pole,  be  a  fixed  point,  and 
let  OM,  called  the  initial  line,  be  a  fixed  line.  Take  P  any  point 
in  the  plane  and  draw  OP.  Denote  OP  by  r  and  the  angle 
MOP  by  0.  Then  r  and  6  are  called  the  polar  coordinates  of 
the  point  P  (r,  #),  and  when  given  will  completely  determine  P. 

For  example,  the  point  (2,  15°)  is  plotted  by  laying  off  the 
angle  MOP  —  lb°  and  measuring  OP  =  2. 

OP,  or  r,  is  called  the  radius  vector,  and  9 
the  vectorial  angle,  of  P.  These  quantities  may 
be  either  positive  or  negative.  A  negative 
value  of  6  is  laid  off  in  the  direction  of  the 
motion  of  the  hands  of  a  clock,  a  positive 
angle  in  the  opposite  direction.  After  the 
angle  6  has  been  constructed,  positive  values  of  r  are  measured 
from  0  along  the  terminal  line  of  6,  and  negative  values  of  r 
from  0  along  the  backward  extension  of  the  terminal  line.  It 
follows  that  the  same  point  may  have  more  than  one  pair  of 
coordinates.  Thus  (2,  195°),  (2,  -165°),  (-2,  15°),  and 
(—  2,  —  345°)  refer  to  the  same  point.  In  practice  it  is  usually 
convenient  to  restrict  6  to  positive  values. 

Plotting  in  polar  coordinates  is  facilitated  by  using  paper  ruled 
as  in  figs.  100  and  101.  The  angle  6  is  determined  from  the  num- 
bers at  the  ends  of  the  straight  lines,  and  the  value  of  r  is  counted 
off  on  the  concentric  circles,  either  towards  or  away  from  the 
number  which  indicates  6,  according  as  r  is  positive  or  negative. 

When  an  equation  is  given  in  polar  coordinates,  the  corre- 
sponding  curve    may  be   plotted    by  giving   to   6   convenient 


POLAR  coordinates 


119 


values,  computing  the  corresponding  values  of  r,  plotting  the 
resulting  points,  and  drawing  a  curve  through  them. 

Ex.  1.  r  =  a  cos  6. 

a  is  a  constant  which 
may  be  given  any  con- 
venient value.  We  may 
then  find  from  a  table  oi 
natural  cosines  the  value 
of  r  which  corresponds 
to  any  value  of  6.  By 
plottiug  the  points  corre- 
sponding to  values  of  6 
from  0°  to  90°,  we  obtain 
the  arc  ABCO  (fig.  100). 
Values  of  6  from  90°  to 
180°  give  the  arc  ODEA. 
Values  of  0  from  180° 
to  270°  give  again  the 
arc  ABCO,    and    those 

from  270°  to  360°  give  the  arc  ODEA.   Values  of  6  greater  than  360  'can 
clearly  give  no  points  not  already  found.    The  curve  is  a  circle  (§  63). 


Ex.  2.  r  =  a  sin  3  0. 

As  6  increases  from  0°  to 
30°,  r  increases  from  0  to  a; 
as  0  increases  from  30°  to  60  , 
r  decreases  from  a  to  0  ;  the 
point  (?•,  6)  traces  out  the 
Loop  0A0  (fig.  101).  As  6 
increases  from  G0°  to  90°, 
r  is  negative  and  decreases 
from  0  to  —  a  ;  as  6  increases 
from  90°  to  120°,  r increases 
from  —a  to  0;  the  point 
(r,  6)  traces  out  the  loop 
OB 0.  As  8  increases  from 
120°  to  180°,  the  point  (r,  6) 
traces  out  the  loop  OCO. 
Larger  values  of  $  give 
points  already  found,  since 
sin  3  (180° +  '(9)  =  -  sin  3  ft 
sin  3  (G0°  +  6)  =  —  sin  3  6. 


The    three    loops    are    congruent,    because 
This  curve  is  called  a  rose  of  three  leaves. 


120 


POLAR  COORDINATES 


Ex.  3.  r2  =  2  rt2  cos  2  0. 

Solving  for  r,  we  have 


±aV2cos2  0. 

Hence,  corresponding  to  any  values  of  0  which  make  cos  26  positive,  there 
will  be  two  values  of  r  numerically  equal  and  opposite  in  sign  and  two 
corresponding  points  of  the  curve  symmetrically  situated  with  respect  to 
the  pole.  If  values  are  assigned  to  6  which  make  cos  2  9  negative,  the  cor- 
responding values  of  r  will  be  imaginary  and  there  will  be  no  points  on 
the  curve. 

Accordingly,  as  6  increases 
from  0°  to  45°,  r  decreases  nu- 
merically from  a  to  0,  and  the 
portions  of  the  curve  in  the  first 
and  the  third  quadrant  are  con- 
structed ;  as  0  increases  from  45° 
to  135°,  cos  2  6  is  negative,  and 
there  is  no  portion  of  the  curve  between  the  lines  6  =  45°  and  6  =  135°; 
finally,  as  6  increases  from  135°  to  180°,  r  increases  numerically  from  0 
to  a,  and  the  portions  of  the  curve  in  the  second  and  the  fourth  quadrant 
are  constructed.  The  curve  is  now  complete,  as  we  should  only  repeat  the 
curve  already  found  if  we  assigned  further  values  to  6;  it  is  called  the 
lemniscate  (fig.  102). 

61.  The  spirals.  Polar  coordinates  are  particularly  well 
adapted  to  represent  certain  curves  called  spirals,  of  which 
the   more  important  follow: 

Ex.  1.    The  spiral  of  Archimedes, 


Fig.   102 


In  plotting,  6  is  usually  considered 
in  circular  measure.  When  9  =  0, 
r  —  0,  and  as  6  increases,  r  increases, 
so  that  the  curve  winds  infinitely  often 
around  the  origin  while  receding  from 
it  (fig.  103).    In  the  figure  the  heavy 

line  represents  the  portion  of  the  spiral  corresponding  to  positive  values 
of  6,  and  the  dotted  line  the  portion  corresponding  to  negative  values  of  6. 

Ex.  2.    The  hyperbolic  spiral, 


Fig.  103 


As  6  increases  indefinitely,  r  approaches  zero.    Hence  the  spiral  winds 
infinitely  often  around  the  origin,  continually  approaching  it  but  never 


THE  STRAIGHT  LIXE 


121 


reaching  it  (fig.  104).    As  6  approaches  zero,  r  increases  without  limit. 
If  P  is  a  point  on  the  spiral  and  NP  is  the  perpendicular  to  the  initial  line. 


NP 


6 


Fig.  104 


Hence,  as  6  approaches  zero  as 
a  limit,  NP  approaches  a  (§  95). 
Therefore  the  curve  comes  con- 
stantly nearer  to,  but  never  reaches, 
the  line  LK,  parallel  to  OM  at  a 
distance  a  units  from  it.  This  line  is  therefore  an  asymptote.  In  the 
figure  the  dotted  portion  of  the  curve  corresponds  to  negative  values  of  6. 

Ex.  3.   The  logarithmic  spiral, 
r  =  &*. 

When  6  =  0,  r  =  1.  As  9  increases,  r 
increases,  and  the  curve  winds  around 
the  origin  at  increasing  distances  from 
it  (fig.  10")).  "When  6  is  negative  and 
increasing  numerically  without  limit,  r 
approaches  zero.  Hence  the  curve  winds 
infinitely  often  around  the  origin,  continu- 
ally approaching  it.  The  dotted  line  in  the 
figure  corresponds  to  negative  values  of  6. 

A  property  of  this  spiral  is  that  it  cuts 
the  radius  vectors  at  a  constant  angle.  The 
student  may  prove  this  after  reading  §  103. 

We  shall  now  give  examples  of  the  derivation  of  the  polar 
equation  of  a  curve  from  the  definition  of  the  curve. 

62.  The  straight  line.    Let  LK  (fig.  106)  be  a  straight  line 
perpendicular  to  OD.     Let  the   angle  MOD  be   denoted  by  a, 
and  let  OD  =  p  ;  then  p  is  the  normal 
distance  of  LK  from  the  pole. 

Let  P(r,  6}  be   any  point  of  LK. 
Then,  by  trigonometry, 


Fig.   105 


OP  cos  D  OP 
r  cos  (0  —  a) 


OD, 


(1) 


which  is  the  equation  of  the  straight 
line. 


122  POLAR  COORDINATES 

I  f  a  =  0  and  p  —  a,  we  have  the  special  equation 
r  cos  6  =  a, 
or  r  =  a  sec  6.  (2) 

If  the  straight  line  passes  through  the  origin,  p  =  0.  The 
equation  of  the  line  then  becomes 

cos  (6  —  a)  =  0, 
or  simply  6  —  —  +  a, 

which  is  of  the  form  6  =  c.  (3) 

63.  The  circle.  Let  C(b,  a)  be  the  center  and  a  the  radius  of 
a  circle  (fig.  107).  Let  P(r,  0)  be  any  point  of  the  circle,  and 
draw  the  straight  lines  OC,  OP,  and  CP. 

By  trigonometry,  we  have 

OP2  +  OC2  -  2  OP  •  OC  cos  POC=CP\ 

Noting  that  cos  POC  =  cos  (0  —  a),  OP  =  r,  OC  =  h,  and 
CP  =  «,  and  substituting  in  the  equation,  we  have  the  result 

r2-2rb  cos  (6  -  a)  +  62  =  a2    (1) 

as  the  polar  equation  of  the  circle. 

When  the  origin  is  at  the  center  of 
the  circle,  b  =  0  and  (1)  becomes  simply 

Fig.  107 

When  the  origin  is  on  the  circle,  b  =  a  and  (1)  becomes 

r-2«cos(6>-«)=0; 
which  may  be  written    r  =  aQ  cos  6  +  ay  sin  0,  (3) 

where  a0  and  ax  are  the  intercepts  on  the  lines  6=0  and  6  =  — 
respectively. 

When  the  origin  is  on  the  circle  and  the  initial  line  is  a 
diameter,  (3)  becomes  ?.  _  a  CQg  ^  ^ 

When  the  origin  is  on  the  circle  and  the  initial  line  is  tangent 

to  the  circle,  (3)  becomes  .    n  ^rx 

r  =  a  sin  0.  (5) 


THE  LIMAgON 


123 


64.  The  limaQon.  TJirough  any  fixed  point  0  (fig.  108)  on  the 
circumference  of  a  fixed  circle  draw  any  line  cutting  the  circle 
again  at  D,  and  lay  off  on  this  line  a 
constant  length  measured  from  D  in 
either  direction.  The  locus  of  the 
points  P  and  Q  thus  found  is  a  curve 
called  the  limagon. 

Take  0  as  the  pole,  and  the  diameter 
OA  as  the  initial  line,  of  a  system  of 
polar  coordinates,  and  call  the  diame- 
ter of  the  circle  a  and  the  constant 
length  b.  Then  it  is  clear  that  the 
entire  locus  can  be  found  by  caus- 
ing OD  to  revolve  through  an  angle  of  3»>0°  and  laying  off 
Dl'=b,  always  in  the  direction  of  the  terminal  line  of  AOD. 

Let  P  be  (>•,  0),  where  6  =  AOD.  Then  r  =  OD+DP  when 
6  is  in  the  first  or  the  fourth  quadrant,  and  r=—  (>I)  +  DP 
when  6  is  in  the*  second  or  the  third 
quadrant.  But  it  appears  from  the 
figure  that  OD  =  OA  cos  6  when  6  is 
in  the  first  or  the  fourth  quadrant, 
and  that  OD  =  —  OA  cos  6  when  0  is 
in  the  second  or  the  third  quadrant. 
Hence,  for  any  point  on  the  limac.on, 
r  =  a  cos  6  +  b. 


-,„,'(-;,) 


-ct>»:'(-af) 


Fn;.   109 


In  studying  the  shape  of  the 
curve  there  are  three  cases  to  be 
distinguished : 

1.  b>a.    r  is  always  positive;  the  curve  appears  as  in  fig.  108. 

2.  b  <  a.    r   is   positive    when    cos  6  > •>    negative    when 

h  1  ^ 

c,osd< ,  and  zero  when  cos  0  = The  curve  appears  as 

a  a 

in  fig.  109. 

3.  b  =  a.   The  equation  now  becomes 

Q 

r  —  a  (cos  6  + 1)  =  2  a  cos2  -  • 


124 


POLAR  COORDINATES 


Here  r  is  positive,  except  that  when  0  =  180°  r  is  zero.    The 
curve  appears  as  in  fig.  110   and  is  called  the   cardioid. 

The  cardioid  is  an  epicycloid  for 
which  the  radius  of  the  fixed  circle 
equals  that  of  the  rolling  circle.  The 
proof  of  this  is  left  to  the  student. 

65.  Relation  between  rectangular 
and  polar  coordinates.  Let  the  pole 
0  and  the  initial  line  OM  of  a  sys- 
tem of  polar  coordinates  be  at  the 
same  time  the  origin  and  the  axis 
of  re  of  a  system  of  rectangular 
coordinates.      Let   P  (fig.    HI)    be  FlG    110 

any  point  of   the  plane,    (x,  y)  its 

rectangular    coordinates,    and    (r,    0)    its    polar    coordinates. 
Then,    by   the    definition    of   the   trigonometric    functions, 


cos  0 


sin  6 


(1) 


Whence  follows,  on  the  one  hand 
x  =  r  cos  0, 
y  —  r  sin  6  ; 
and,  on  the  other  hand, 

r=Vx2+y2,         si?i/}  — 


Fig.  Ill 


COS0 


(2) 


Vx2+y2  V;r'2+u 

By  means  of  (1)  a  transformation  can  be  made  from  rectangular 
to  polar  coordinates,  and  by  means  of  (2)  from  polar  to  rectangular 
coordinates. 


Ex.  1.    The  equation  of  the  cissoid  (§  48)  is 
x3 

y  =  7T-. — z  • 


Substituting  from  (1)  and  making  simple  reductions,  we  have  the  polar 
equation  . 

r  = ^ — 


THE  CONIC  125 

Ex.  2.    The  polar  equation  of  the'  lemniscate  (Ex.  3,  §  60)  is 
7-2=2a2cos2  0. 

Placing  cos  26  =  cos2 6—  sin20  and  substituting  from  (2),  we  have  the 
rectangular  equation 

(x2  +  r)2=2«2(2-2-r)- 

66.  The  conic,  the  focus  being  the  pole.  From  §  46,  the  equa- 
tion of  a  conic  when  the  axis  of  x  is  an  axis  of  the  conic  and 
the  axis  of  y.  is  a  directrix  is 

We  may  transfer  to  new  axes  having  the  focus  of  the  conic  as 
the  origin  and  the  axis  of  the  conic  as  the  axis  of  x  by  placing 

x  =  c  +  x',       y  =  y\ 

thus  obtaining  x' '2  +  y' a  =  f -( ./•'  +  rf. 

If  we  now  take  a  system  of  polar  coordinates  having  the  focus 
as  the  pole  and  the  axis  of  the  conic  as  the  initial  line,  we  have 

xr=rco8  0,         y,=  ram0. 

The  equation  then  becomes 

r9=ea(rcos^  +  c)9, 

which  is  equivalent  to  the  two  equations 

ce                                     ce 
r  — »  r  = • 

1  —  e  cos  6  1  +  >■  cos  6 

Either  of  these  equations  alone  will  give  the  entire   conic. 

To  see  this,  place  6  =  dl  in  the  second  equation,  obtaining 

—  ce 

r  = 

1      1  +  e  cos  0X 

Now  place  6  =  it  +  6X  in  the  first  equation,  obtaining  r  =  —  r{. 

The  points  (r^  6^)  and  (—  r^  tt  +  #x)  are  the  same.    Hence  any 

point  which  can  be  found  from   the   second   equation   can    be 

found  from  the  first. 

I  hereiore  r  = 

1  —  f  cos  6 

is  the  required  polar  equation. 


120 


POLAR  COORDINATES 


67.  Examples.  Polar  coordinates  may  be  used  with  great 
advantage  in  the  solution  of  problems  involving  a  number  of 
straight  lines  radiating  from  a  given  point,  the  given  point 
then  being  taken  as  the  pole  of  the  system  of  coordinates. 
This  use  is  illustrated  in  the  following  examples: 

Ex.  1.  Prove  that  if  a  secant  is  drawn 
through  the  focus  of  a  conic,  the  sum  of  the 
reciprocals  of  the  segments  made  by  the 
focus  is  constant. 

Let  P1P2  (fig.  112)  be  any  secant  through 
the  focus  F,  and  let  FP1  =  rv  FP2  =  r2,  and 
the  angle  MFP1  =  0.  Then  the  polar  coordi- 
nates of  P1  are  (rv  6)  and  those  of  P2  are 
(r2,  6  +  tt).  From  the  polar  equation  of  the 
conic,  we  have 

r.  — £! ,  Fig.  112 


Hence 


I  +  I 


1  —  e  cos  (6  +  7r)       1  +  e  cos  I 
2 


Ex.  2.  Find  the  locus  of  the  middle 
of  a  circle   all  of  which  pass  through   a 

Take  any  circle  with  the  center  C 
(fig.  113),  and  let  0  be  any  point  in  the 
plane.  If  0  is  taken  for  the  pole,  and  OC 
for  the  initial  line,  of  a  system  of  polar 
coordinates,  the  equation  of  the  circle  is 

r2  -  2  rb  cos  d  +  i2  -  a2  =  0.         (1) 

Let  PXP2  be  any  chord  through  O,  and 
let  OP1  =  rv  OP2  =  r2.  Then  1\  and  r2  are 
the  two  roots  of  equation  (1)  which  corre- 
spond to  the  same  value  of  0.    Hence 

?•,  +  r„  =  2  b  cos  6. 


points  of   a  system  of  chords 
fixed  point. 


If  Q  is  the  middle  point  of  PXP2  and  we  now  place  OQ  —  r,  we  have 


ri  + 


b  cos  0. 


But  this  is  the  polar  equation  of  a  circle  through  the  points  0  and  C. 


Plot  the  following  curves 

1.  r  =  a  sin  2  0. 

2.  r  =  a  cos  3  0. 


3.  r  =  a  sin-' 
Z 


4.  r  =  a  cos 


5.  r 


6.  r8  =  a2  sin  0. 

7.  j-2  =  «°-  sin  3  0. 

8.  r2  =  a2  sin  4  0. 

9.  >•=  «(l  +  sin0). 

10.  r=  a  (2  +  sin0). 

11.  r=  «  (1  +  cos  2  0), 

12.  r=  a  (1  —  cos  2  0). 

13.  /■  =  a(l+  cos  3  0). 

14.  r  =  a  (2  +  cos  2  0). 

15.  r  =  a(l  +  2sin0). 

16.  r  =  a(l  +  2  cos  2  0). 

17.  >•  =  «(! +  2  cos  3  0). 


PROBLEMS 

18. 

o   ,     •    30 
r  =  2  +  sin  —  • 

19. 

r2=«2(l-COS0). 

20. 

r  =  «2(l  +  2cos20) 

21. 

r  =  a  tan  0. 

22. 

/•  =  a  tan  2  0. 

23. 

,      0 

;•  =  a  tan  -  • 

24. 

r  =  a  sec  2  0. 

25. 

0 

/■  =  a  sec      • 

127 


26.  ?•  =  a  sec2-- 

27.  r  =  a(l  +  scc0). 

28.  r  =  a(l  +  2  sec  0). 

29.  r=  «(2  +  sec0). 

30.  /•  cos  0  =  a  cos  2  0. 


31.  r  = 

32.  r  =  1 

33.  >•  =  - 


COS0       sill  0 
2  0. 


Plot  each  pair  of  the  following  curves  in  one  diagram  and  find 
their  points  of  intersection  : 


34.  rcos 


35.   r cos 


36.  r  cos (d  —  ~)  =a  V2,  r  =  2  a  cos  0. 

37.  r*  =  aasin0,  r*=a3sin20. 

38.  r  =  a(l  +  sin  2  0),  r*  =  4  a2  sin  2  0. 

39.  >-2  =  a2  sin  0,   z-2  =  a2  sin  3  0. 


128  POLAR  COORDINATES 

40.  0  is  a  fixed  point  and  LK  a  fixed  straight  line.  If  any  straight 
line  through  0  intersects  LK  in  Q,  and  a  point  P  is  taken  on  this 
line  so  that  OP  ■  OQ  =  A-2,  find  the  locus  of  P. 

41.  A  straight  line  OA  of  constant  length  a  revolves  about  0. 
From  A  a  perpendicular  is  drawn  to  a  fixed  straight  line  OM,  inter- 
secting it  in  B.  From  B  a  perpendicular  is  drawn  to  OA,  intersecting 
it  in  P.    Find  the  locus  of  P,  OM  being  taken  as  the  initial  line. 

42.  0  is  a  fixed  point  of  a  circle  of  radius  a,  and  OM  is  a  fixed 
straight  line  passing  through  the  center  of  the  circle.  A  straight 
line  is  drawn  from  O  to  any  point  Px  of  the  circle,  and  from  P1  a 
straight  line  is  drawn  perpendicular  to  OM,  meeting  OM  at  Q.  From 
Q  a  straight  line  is  drawn  perpendicular  to  OPv  meeting  OP1  at  P. 
Find  the  equation  of  the  locus  of  P,  taking  O  as  the  origin  of  coor- 
dinates and  OM  as  the  initial  line. 

43.  MN  is  a  straight  line  perpendicular  to  the  initial  line  at  a 
distance  a  from  0.  From  0  a  straight  line  is  drawn  to  any  point  B 
of  MN.  From  B  a  straight  line  is  drawn  perpendicular  to  OB,  inter- 
secting the  initial  line  at  C.  From  C  a  straight  line  is  drawn  per- 
pendicular to  BC,  intersecting  MN  at  D.  Finally,  from  D  a  straight 
line  is  drawn  perpendicular  to-  CD,  intersecting  OB  at  P.  Find 
the  locus  of  P. 

Transform  the  following  equations  to  polar  coordinates  : 

44.  xy  =  7.  46.  xi  +  x2f  —  ah?  =  0. 

45.  x2  +  f  -  8  ax  -  Say  =  0.  47.   (.«2  +  iff  =  d2(x2  -  if). 

48.  Find  the  polar  equation  of  the  strophoid  when  the  pole  is  0 
and  the  initial  line  is  OA  (fig.  82). 

Transform  the  following  equations  to  rectangular  coordinates  : 

49.  r  cos  (o-7^)  +  r  cos  (d +  ^)  =  12. 

50.  r  =  «  sin  0.  51.  r  =  a  tan  6. 

52.  Find  the  Cartesian  equation  of  the  rose  of  four  petals 
r  =  a  sin  20. 

53.  Find  the  Cartesian  equation  of  the  cardioid  r  =  a(l—  cos  0). 

54.  Find  the  Cartesian  equation  of  the  limacon  r  =  a  cos  6  +  b. 


PROBLEMS  129 

55.  In  a  parabola  prove  that  the  length  of  a  focal  chord  which 
makes  an  angle  of  30°  with  the  axis  of  the  curve  is  four  times  the 
focal  chord  perpendicular  to  the  axis. 

56.  A  comet  is  moving  in  a  parabolic  orbit  around  the  sun  at  the 
focus  of  the  parabola.  When  the  comet  is  100,000,000  miles  from 
the  sun  the  radius  vector  makes  an  angle  of  60°  with  the  axis  of  the 
orbit.  What  is  the  equation  of  the  comet's  orbit  ?  How  near  does 
it  come  to  the  sun  ? 

57.  A  comet  moving  in  a  parabolic  orbit  around  the  sun  is 
observed  at  two  points  of  its  path,  its  focal  distances  being  5  and  15 
million  miles,  and  the  angle  between  them  being  90°.  How  near 
does  it  come  to  the  sun  ? 

58.  If  a  straight  line  drawn  through  the  focus  of  an  hyperbola 
parallel  to  an  asymptote  meets  the  curve  at  P,  prove  that  FP  is  one 
fourth  the  chord  through  the  focus  perpendicular  to  the  transverse 
axis. 

59.  The  focal  radii  of  a  parabola  are  extended  beyond  the  curve 
until  their  lengths  are  doubled.    Find  the  locus  of  their  extremities. 

60.  If  1\  and  P2  are  the  points  of  intersection  of  a  straight  line 
drawn  from  any  point  0  to  a  circle,  prove  that  <>/\  •  OPa  is  constant. 

61.  If  Pj  and  P2  are  the  points  of  intersection  of  a  straight  line 
from  any  point  O  to  a  fixed  circle,  and  Q  is  a  point  on  the  same 

L'  OP  ■  OP 

straight  line  such  that  (>(}  =  —   — L ^>  find  the  locus  of  (). 

oi\  +  <>P, 

62.  Secant  lines  of  a  circle  are  drawn  from  the  same  point  on  the 
circle,  and  on  each  secant  a  point  is  taken  outside  the  circle  at  a 
distance  equal  to  the  portion  of  the  secant  included  in  the  circle. 
Find  the  locus  of  these  points. 

63.  From  a  point  O  a  straight  line  is  drawn  intersecting  a  fixed 
circle  at  P,  and  on  this  line  a  point  Q  is  taken  so  that  OP  •  OQ  =  k2. 
Find  the  locus  of  Q. 

64.  Find  the  locus  of  the  middle  points  of  the  focal  chords  of 
a  conic. 

65.  Find  the  locus  of  the  middle  points  of  the  focal  radii  of 
a  conic. 

66.  If  P1FP2  and  QXFQ2  are  two  perpendicular  focal  chords  of  a 

conic,  prove  that  — h —  is  constant. 

'  *  PXF  •  FP2  T  Qf  •  FQ2 


CHAPTER  IX 
SLOPES  AND  AREAS 

68.  Limits.  A  variable  is  said  to  approach  a  constant  as  a 
limit,  when,  under  the  laiv  which  governs  the  change  of  value  of 
the  variable,  the  difference  between  the  variable  and  the  constant 
becomes  and  remains  less  than  any  quantity  which  can  be  named, 
no  matter  how  small. 

If  the  variable  is  independent,  it  may  be  made  to  approach  a 
limit  by  assigning  to  it  arbitrarily  a  succession  of  values  follow- 
ing some  known  law.    Thus,  if  x  is  given  in  succession  the  values 

13  7  2n-l 

*x=2'  x*=r  x*=r  •••'  x^-¥~' 

and  so  on  indefinitely,  it  approaches  1  as  a  limit.  For  we 
may  make  x  differ  from  1  by  as  little  as  we  please  by  taking  n 
sufficiently  great ;  and  for  all  i  i      7  « 

larger  values  of  n  the  differ-     1 + 1  'f  i- 

ence  between  x  and  1  is  still 

smaller.    This  may  be  made 

evident    graphically   by   marking   off    on    a    number    scale   the 

successive  values   of  x  (fig.  114),  when  it  will  be   seen  that 


Fig.  115 

the  difference   between    x   and   1    soon   becomes    and  remains 
too  minute  to  be  represented. 

Similarly,  if  we  assign  to  x  the  succession  of  values 

1111  ,       1V-       1 


1     2       2         3       3     4       4         5  '     "     v      '      n+1 

x  approaches  0  as  a  limit  (fig.  115). 

130 


LIMITS 


131 


If  the  variable  is  not  independent,  but  is  a  function  of  x,  the 
values  which  it  assumes  as  it  approaches  a  limit  depend  upon 
the  values  arbitrarily  assigned  to  x.  For  example,  let  y  =f(x), 
and  let  x  be  given  a  set  of  values, 


approaching  a  limit  a.    Let  the  corresponding  values  of  y  be 

y*   y*   y8>   y*    •••»   y.»    •••• 

Then,  if  there  exists  a  number  A  such  that  the  difference  between 
y  aifd  A  becomes  and  remains  less  than  any  assigned  quantity,  y 
is  said  to  approach  A  as  a  limit  as  x  approaches  a  in  the  manner 
indicated.  This  may  be  seen 
graphically  in  fig.  116,  where 
the  values  of  x  approaching  a 
are  seen  on  the  axis  of  abscissas, 
and  the  values  of  y  approach- 
ing A  are  seen  on  the  axis 
of  ordinates.  The  curve  of  the 
function  is  continually  nearer 
to  the  line  y  =  A. 

In  the  most  common  cases  the 
limit  of  the  function  depends 
only  upon  the  limit  a  of  the 
independent  variable  and  not 
upon  the  particular  succession 
approaching  a.    This  is  clearly 


of    values    that 
the   case   if  the 


x    assumes   in 
graph   of   the 


function  is  as  drawn  in  fig.  116. 

Ex.  1.    Consider  the  function 

x-  +  3  x  -  4 
y=        x-l        ' 

and  let  x  approach  1  by  passing  through  the  succession  of  values 

x  =  1.1,     !F  =  1.01,     x  =  1.001,     a:  =  1.0001,      •••. 

Then  y  takes  in  succession  the  values 

y  =  5.1,     y  =  5.01,     y  =  5.001,     y  -  5.0001. 

It  appears  as  if  y  were  approaching  the  limit  5.    To  verify  this  we  place 
X  =  1  +  h,  where  It  is  not  zero.    By  substituting  and  dividing  by  h,  we  find 


132  SLOPES  AND  AREAS 

y  =  5  +  h.  From  this  it  appears  that  y  can  be  made  as  near  5  as  we 
please  by  taking  h  sufficiently  small,  and  that  for  smaller  values  of  h, 
y  is  still  nearer  5.  Hence  5  is  the  limit  of  y  as  x  approaches  1.  More- 
over, it  appears  that  this  limit  is  independent  of  the  succession  of  values 
which  x  assumes  in  approaching  1. 

Ex.  2.    Consider  y  = =  as  x  approaches  zero. 

1  -  Vl  -  x 

Give  x  in  succession  the  values  .1,  .01,  .001,  .0001,  •  •  •.  Then  y  takes  the 
values  1.9487, 1.9950,  1.9995, 1.9999,  •  •  •,  suggesting  the  limit  2. 

In  fact,  by  multiplying  both  terms  of — =  by  1  +  Vl  —  x,  we  find 

1  -  VI  -  x 
y  =  1  +  Vl—  x  for  all  values  of  x  except  zero. 

Hence  it  appears  that  y  approaches  2  as  a;  approaches  0. 

We  shall  use  the  symbol  =  to  mean  "approaches  as  a  limit." 
Then  the  expressions 

Lima-  =  a 

and  x  =  a 

have  the  same  significance. 

The  expression  ~Limf(x')  =  A 

x  =  a 

is  read  "  The  limit  of  f(x),  as  x  approaches  «,  is  Ay 

69.  Theorems  on  limits.  In  operations  with  limits  the  follow- 
ing propositions  are  of  importance : 

J  1.  The  limit  of  the  sum  of  a  finite  number  of  variables  is  equal 
to  the  sum  of  the  limits  of  the  variables. 

We  will  prove  the  theorem  for  three  variables ;  the  proof  is 
easily  extended  to  any  number  of  variables. 

Let  X,  F,  and  Z  be  three  variables,  such  that  Lim  X=A, 
Lim  Y=B,  him  Z=C.  From  the  definition  of  limit  (§68)  we 
may  write  X  =  A+a,  Y  =  B+b,  Z=C+c,  where  a,  b,  and  c 
are  three  quantities  each  of  which  becomes  and  remains 
numerically  less  than  any  assigned  quantity  as  the  variables 
approach  their  limits. 

Adding,  we  have 

A'+  F+  Z  =  A  +  B  +  C+  a  +  b  +  c. 


LIMITS  133 

Now  if  e  is  any  assigned  quantity,  however  small,  we  may 

make  a,  b,  and  c  each  numerically  less  than  - ,  so  that  a  +  b  +  c  is 

o 

numerically  less  than  e.    Then  the  difference  between  X+Y+  Z 

and  A+B  +  C  becomes  and  remains  less  than  e;  that  is, 

Lim  (A'  +  Y+  Z)=A+B  +  C  =  Lim X+  Lim  Y+  Lim  Z. 

V 
2.  The  limit  of  the  product  of  a  finite  number  of  variables  is 

equal  to  the  product  of  the  limits  of  the  variables. 

Consider  first  two  variables  A'  and  F,  such  that  Lim  X=  A  ami 

Lim  Y=  B.    As  before,  we  have  A=  A  +  a  and  Y=B  +  b.    Hence 

XY=AB+bA+aB+ab. 

Now  we  may  make  a  and  b  so  small  that  bA,  aB,  and  ab  are 

each  less  than  -,  where  e  is  any  assigned  quantity,  no  matter  how 
small.   Hence 

Lim  XY=  AB  =  (Lim  A')  (Lim  F). 

Consider  now  three  variables  A,  F,  Z.  Place  XY  =  U.  Then, 
as  just  proved,       LimPZ=(LimP)(LimZ). 

that  is,  LimA]'Z  =  (LimA)')(LiinZ) 

=  (LiinA)(Lim)')(LimZ). 

Similarly,  the  theorem  may  be  proved  for  any  finite  number 
of  variables. 

v/      3.  The  limit  of  a  constant  multiplied  by  u  variable  is  equal  to 

the  constant  multiplied  by  the  limit  if  the  variable. 

The  proof  is  left  for  the  student. 

4.  The  limit  of  the  quotient  if  two  variables  is  equal  t*>  the 
quotient  of  the  limits  of  the  variables,  provided  the  limit  of  the 
divisor  is  not  zero. 

Let  A'  and  Y  be  two  variables,  such  that  Lim  .V=  J  and 
Lim  Y==B.    Then,  as  before,  X=A+a,  Y=B  +  b. 

X     A+a  t     A'     A      A  +  a      A      aB  —  bA 

IIence  y=i^i,'  mi  y-jriiTi-^^M- 


134 


SLOPES  AND  APEAS 


Now  the  fraction  on  the  right  of  this  equation  may  be  made  less 
than  any  assigned  quantity  by  taking  a  and  b  sufficiently  small. 


Hence 


Lim 


LimX 
Lim  Y ' 


The  proof  assumes  that  B  is  not  zero. 

70.  Slope  of  a  curve.  By  means  of  the  conception  of  a  limit 
we  may  extend  the  definition  of  "  slope,"  given  in  §  6  for  a 
straight  line,  so  that  it  may  be  applied  to  any  curve.  Let 
Px  and  P2  be  any  two  points  upon  a  curve 
(fig.  117).  If  i?  and  P2  are  connected  by 
a  straight  line,   the   slope   of  this  line   is 


y.-yi 


If  P2  and  Px  are  close  enough  to- 


Fig.  117 


gether,  the  straight  line  P,P2  will  differ  only 

a  little  from  the  arc  of  the  curve,  and  its 

slope  may  be  taken  as  approximately  the 

slope  of  the  curve  at  the  point  Pv    Now 

this  approximation  is  closer,  the  nearer  the  point  jP  is  to  Pv 

Hence  we   are   led  naturally  to  the  following   definition: 

The  slope  of  a  curve  at  a  point  Px  (x^  y^)  is  the  limit  approached 


by  the  fraction 


y*-Vx 


ivhere  xn  and  y    are  the  coordinates  of  a 


second  point  P2  on  the  curve  and  where  the  limit  is  taken  as  P^ 
moves  toward  i?  along  the  curve. 

Ex.  1.    Consider  the  curve  y  =  x2  and  the  point  (5,  25)  upon  it,  and  let 

«i  =  5.  !lx  =  25- 

We  take  in  succession  various  values  for  x2  and  y„,  corresponding  to 
points  on  the  curve  which  are  nearer  and  nearer  to  (xv  y^),  and  arrange 
our  results  in  a  table  as  follows : 


xa 

?/2 

x2  —  xx 

y.,  -  vx 

y-2  -  V\ 

x2  —  xx 

G 

30 

1 

n 

11 

5.1 

20.01 

.1 

1.01 

10.1 

5.01 

25.1001 

.01 

.1001 

10.01 

5.001 

25.010001 

.001 

.010001 

10.001 

INCREMENT  135 

The   arithmetical  work  suggests  the  limit  10.    To  verify  this,  place 

ar2  =  5  +  h.    Then  y2  =  25  + 10  h  +  h2.    Consequently  '  2  _  '  *  =  10  +  A,  and 

y«  —  iii  2      l 

as  x„  approaches  x,,  h  approaches  0  and    "  _       approaches  10.    Hence  the 

xi      x\ 

slope  of  the  curve  y  —  x'2  at  the  point  (5,  25)  is  10. 

Ex.  2.    Find  the  slope  of  the  curve  y  =  -  at  the  point  (3,  ^). 

"We  have  here  xx  —  3,  yx  =  \. 

We  place  x„  —  3  +  A,  »/o  = 

3  +  A 

rrv  7.  ~   ''  1      'A.  —    '/l  1 

1  hen     x„  —  x,  =  «,  ?/.,  —  ?/,  = >  aud  — -■ 


9  +  3A  x8  —  z2         9  +  3* 

As  P2  approaches  1\  along  the  curve,  h  approaches  0,  and  the  limit  of 
-  is  —  ;t>  hence  the  slope  of  the  curve  at  the  point  (:5,  .',)  is  —  1. 


In  a  similar  manner  we  may  find  the  slope  of  any  curve  the 
equation  of  which  is  not  too  complicated ;  but  when  the  equa- 
tion is  complicated,  there   is   need  of   a  more  powerful  method 

y  —  y 

for  finding  the  limit  of  — •    This  method  is  furnished  by 

xa  -  xx 
the  operation  known   as  differentiation,  the  first  principles  of 
which  are  explained  in  the  following  articles. 

71.  Increment.  When  a  variable  changes  its  value,  the  quan- 
tity which  is  added  to  its  first  value  to  obtain  its  last  value  is 
called  its  increment.  Thus,  if  x  changes  from  5  to  5|-,  its  incre- 
ment is  i.  If  it  changes  from  5  to  4;j,  the  increment  is  —  \. 
So,  in  general,  if  x  changes  from  rx  to  .>,,  the  increment  is  * 
x2  —  Xj.  It  is  customary  to  denote  an  increment  by  the  symbol 
A  (Greek  delta),  so  that 

A.r  =  x2  —  x^     and     x„  =  xx  +  A#. 

If  y  is  a  function  of  x,  any  increment  added  to  x  will  cause 
a  corresponding  increment  of  y.  Thus,  let  y  =f(x)  and  let  x 
change  from  x1  to  xn.    Then  y  changes  from  yx  to  y2,  where 

y=f(xr)     and     y2  =/(,•)  =/(.rl  +  A*). 
Hence  Ay  =  y„-  yx  =f(xl  +  A.r)  -f(x^. 


136  SLOPES  AND  AKEAS 

72.  Continuity.  A  function  y  is  called  a  continuous  function 
of  a  variable  x  ivhen  the  increment  of  y  approaches  zero  as  the 
increment  of  x  approaches  zero. 

It  is  clear  that  a  continuous  function  cannot  change  its  value 
by  a  sudden  jump,  since  we  can  make  the  change  in  the  function 
as  small  as  we  please  by  taking  the  increment  of  x  sufficiently 
small.  As  a  consequence  of  this,  if  a  continuous  function  has  a 
value  A  when  x=  a,  and  a  value  B 
when  x  =  b,  it  will  assume  any  value 
C,  lying  between  A  and  B,  for  at  V 
least  one  value  of  x  between  a  and  h 
(fig.  118).  V 

In  particular,   if  f(a)   is   positive 
and  f(V)  is  negative,  f(x)  =  0  for  at  FlG    n8 

least  one  value  of  x  between  a  and  b. 

When  Ax  and  Ay  approach  zero  together,  it  usually  happens 

that  —  approaches  a  limit.     In  this  case  y  is  said  to  have  a 

derivative,  defined  in  the  next  article. 

73.  Derivative.  When  y  is  a  continuous  function  of  x,  the  deriva- 
tive of  y  with  respect  to  x  is  the  limit  of  the  ratio  of  the  increment 
of  y  to  the  increment  of  x,  as  the  increment  of  x  approaches  zero. 

The  derivative  is  expressed  by  the  symbol  — ;  or,  if  y  is 

expressed  by  f(x),  the  derivative  may  be  expressed  by  f'(x). 
Thus,  if  y  =f(x), 

dv      j., ,  N      T  .     Ay      T  .      f(x  +  Ax~)  —f(%) 
-f  —f(x\  =  Lim  -£  =  Lim1^ — J- — J-±^-. 

dx  ijioAx      ax-=o  Ax 

The  process  of  finding  the  derivative  is  called  differentiation, 
and  we  are  said  to  differentiate  y  with  respect  to  x.  The  process 
involves,  according  to  the  definition,  the  following  four  steps : 

1.  The  assumption  of  an  increment  of  x. 

2.  The  computation  of  the  corresponding  increment  of  y. 

3.  The  division  of  the  increment  of  y  by  the  increment  of  x. 

4.  The  determination  of  the  limit  approached  by  this  quotient 
as  the  increment  of  x  approaches  zero. 


DIFFERENTIATION  OF  A  POLYNOMIAL  137 

Ex.  1.    Find  the  derivative  of  y  when  y  =  x3. 

(1)  Assume  Ax  =  Ji. 

(2)  Compute  Ay  =  (x  +  /<)3  -  x3  =  3  x2h  +  3  xJr  +  hs. 

(3)  Find  ^  =  3  x2  +  3  xh  +  h2. 

,1, 

(4)  The  limit  is  evidently  3  x2.    Hence  —■  =  3  x2. 

Ex.  2.    Find  the  derivative  of  -  • 

1  x 

(1)  Place  y  =  -  and  assume  Ax  =  //. 

(2)  Compute  Ay  =  ^-i  =  -^A_. 

(3)  Find  ^  =  __1— . 

Ax'  «■  +  ■**    !  rf 

(4)  The  limit  is  cleariy =  i  and  therefore  '-—  = -. 

x"  ax  j- 

It  appears  that  the  operations  of  finding  the  derivative  oif(x) 
are  exactly  those  which  are  used  in  finding  the  slope  of  the  curve 
y=f(x).  Hence  the  derivative  is  a  fund  inn  which  gives  I  In'  slope 
of  the  curve  at  each  point  of  it. 

74.  Differentiation  of  a  polynomial.  The  obtaining  of  a  deriv- 
ative by  carrying  out  the  operations  of  the  last  article  is  too 
tedious  for  practical  use.  It  is  more  convenient  to  use  the 
definition  to  obtain  general  formulas  which  may  be  used  for 
certain  classes  of  functions.  In  this  article  we  shall  derive 
all  formulas  necessary  to  differentiate  a  polynomial, 

1.  — ^- — -=waof-1,  where  n  is  a  positive  integer  and  a  any 

constant. 

Let  y  =  ax". 

(1)  Assume  Ax  =  h. 

(2)  Then  Ay  =  a  (x  +  h)n  —  axn 


-•( 


,-ij  +  "("     1>a»-'/t'+  ...+/<' 


(3)  U«.^.  +  »£gJi-.-.J  +  ...  +  i 

(4)  Taking  the  limit,  we  have  -^  =  nax"  ~\ 


138 


SLOPES  AND  AKEAS 


d  (ax) 


=  a,  where  a  is  a  constant. 


This  is  a  special  case  of  the  preceding  formula,  n  being  here 
equal  to  1.    The  student  may  prove  it  directly. 

dc 
3.  —  =  0,  where  c  is  a  constant. 
d.v 

Since  c  is  a  constant,  Ac  is  always  0,  no  matter  what  the 


value  of  x.    Hence 


Ac 

Ax 


dc 
0,  and  consequently  the  limit  —  =  0. 


4.   The    derivative    of  a  polynomial   is  found   by   adding   the 
derivatives   of  the   terms  in  order. 

This  is  a  special  case  of  a  more  general  theorem  (3,  §  82). 
The  proof  of  the  special  case  before  us  may  be  easily  given 
by  the  student  or  may  be  assumed  temporarily. 
Ex.    Find  the  derivative  of 

f(x)  =  G  x5  -  3  xi  +  5  xs  -  7x2  +  8  x  -  2. 
Applying  formulas  1,  2,  or  3  to  each  term  in  order,  we  have 
f  {x)  =  30  x-4  -  12  x3  +  15  x2  -  14  x  +  8. 

75.  Sign  of  the  derivative.  A  function  of  x  is  called  an 
increasing  function  when  an  increase  in  x  causes  an  increase 
in  the  function.  A  function  of  x  is  called 
a  decreasing  function  when  an  increase  in  x 
causes  a  decrease  in  the  function.  The  graph 
of  a  function  runs  up  toward  the  right  hand 
when  the  function  is  increasing  and  runs 
down  toward  the  right  hand  when  the  func- 
tion is  decreasing.  Thus  x~—  x  —  6  (fig.  119) 
is  decreasing  when  x  <  \  and  increasing  when 
x>\. 

The  sign  of  the  derivative  enables  us  to 
determine  whether  a  function  is  increasing 
or  decreasing  in  accordance  with  the  follow- 
ing theorem : 

When  the  derivative  of  a  function  is  posi- 
tive, the  function  is  increasing;  when  the  derivative  is  negative,  the 
function  is  decreasing. 


SIGN  OF  THE  DEKIYATIVE 


139 


To  prove  this,  consider  y  =f(x),  and   let  us  suppose  that 

—  is  positive.  Then,  since  -~  is  the  limit  of  — ,  it  follows  that  — - 
dx  dx  Ax  Ax 

is  positive  for  sufficiently  small  values  of  Ax;  that  is,  if  Ax 
is  assumed  positive,  Ay  is  also  positive,  and  the   function  is 

increasing.  Similarly,  if  ~  is  negative,  Ay  and  Ax  have  oppo- 
site signs  for  sufficiently  small  values  of  Ax,  and  the  function  is 
decreasing  by  definition. 

Ex.  1.    If  y  =  x1  —  x  —  6,  —  =  2x  —  1,  which  is  negative  when  x  <  k 
dx 
and  positive  when  x>\.    Hence  the  function  is  decreasing  when  x<  \  and 
increasing  when  x>  \,  as  is  shown  in  fig.  119. 


Ex.  2.    If  y=  l(x3-3o;2-9x  +  27): 


|x»-|r-|=|(x+l)(x-3). 


Now  —  is  positive  when  x<  —  1,  negative  when  -KK3,  and  positive 

dx 

when  x  >  3.     Hence  the  function  is  increasing  when  x<  —  1,  decreasing 
when  a:  is  between    —1  and  3,  and  increas- 
ing when  ./•>:;  (fig.  120). 

It  remains  to  examine  the  cases  in 

which  -~  =  0.    Referring  to  the  two  ex- 
dx 

amples  just  given,  we  see  that  in  each 
the  values  of  x  which  make  the  deriv- 
ative zero  separate  those  for  which  the 
function  is  increasing  from  those  for 
which  the  function  is  decreasing.  The 
points  on  the  graph  which  correspond 
to  these  zero  values  of  the  derivative 
can  be  described  as  turning  points. 

Likewise,  whenever  f'(x)  is  a  continuous  function  of  x,  the 
values  of  x  for  which  it  is  positive  are  separated  from  those 
for  which  it  is  negative  by  values  of  x  for  which  it  is  zero 
(§  72).  Now  in  most  cases  which  occur  in  elementary  work, 
f'(x)  is  a  continuous  function.    Hence  we  may  say, 

The  values  of  x  for  which  a  function  changes  from  an  increas- 
ing to  a  decreasing  function  are,  in  general,  values  of  x  which 
make  the  derivative  equal  to  zero. 


Fig.  120 


140 


SLOPES  A^'J)  AREAS 


The  converse  proposition  is,  however,  not  always  true.  A 
value  of  x  for  which  the  derivative  is  zero  is  not  necessarily  a 
value  of  x  for  which  the  function  changes 
from  increasing  to  decreasing  or  from 
decreasing  to  increasing.    For  consider 

l(a;3-9o;2-h27^-19). 

Its  derivative  is  x%—  6  x  +  9  =  (x  —  3)2, 
which  is  always  positive.  The  function 
is  therefore  always  increasing.  When 
x  =  3  the  derivative  is  zero,  and  the 
corresponding  shape  of  the  graph  is 
shown   in   fig.  121. 

76.  Tangent  line.  A  tangent  to  a  curve  is  the  straight  line 
approached  as  a  limit  by  a  secant  line  as  two  points  of  intersection 
of  the  secant  and  the  curve  are  made  to  approach  coincidence. 

Let  Px  and  P2  be  two  points  on  a  curve.  Then  if  a  secant  is 
drawn  through  Px  and  P2  of  a  curve  (fig.  122)  and  the  point 
P2  is  made  to  move  along  the  curve  toward  i^,  which  is  kept 
fixed  in  position,  the  secant  will  turn  on  Px  as  a  pivot  and  will 
approach  as  a  limit  the  tangent  PXT.  The  point  Px  is  called  the 
point  of  contact  of  the  tangent. 

From  the  definition  it  follows  that  the  slope 
of  the  tangent  is  the  same  as  the  slope  of  the 
curve  at  the  point  of  contact ;  for  the  slope 
of  the  tangent  is  evidently  the  limit  of  the 
slope  of  the  secant,  and  this  limit  is  the  slope 
of  the  curve,  by  §  70. 

The  equation  of  the  tangent  is  readily  written  by  means  of 
§  28   when   the   point   of   contact   is   known.     Let    (x^  y^)    be 


Fig.   122 


denote  the  value  of  -~  when 
dx 


the  point  of  contact,  and  let 
x  =  xt  and  y  —  yx.    Then  (xx,  y^)  is  a  point  on  the  tangent  and 
-~\  is  its  slope.    Therefore  its  equation  is 


;).(• 


(1) 


THE  DIFFERENTIAL 


141 


Ex.  1.    Find  the  equation  of  the  tangent  to  the  curve  y  =  x*  at  the 
point  (xv  yx)  on  it. 

Using  formula  (1),  we  have 

y  -  >h  =  :j  %l  (x  -  arx). 

But  since  (xv  yx)  is  on  the  curve,  we  have  y1  =  a;*.    Therefore  the  equa- 
tion can  be  written  ,,     .,         „     3 

y  =  :>  xf  x  —  2  arf.  T 

Ex.  2.    Find  the  equation  of  the  tangent  to 
y  =  x*  +  3  x 
at  the  point  the  abscissa  of  which  is  2. 

dy 


dx 


x  +  :i. 


Ifz^then^lOandg)^?.  ^    m 

Therefore  the  equation  is 

y-  lo  =  7(.#  -2),     or     y-7x  —  4. 

If  ?T  (fig.  123)  is  a  tangent  line  and  <f>  the  angle  it  makes 
with  OX,  its  slope  equals  tan  (/>,  by  ^  -'52.     Hence 

,7„ 


tan  (/>  = 


l.r 


77.  The  differential.  Lot  the  function /(./)  be  represented  by 
the  curve  y=f(.r),  and  let  /'  and  <J  be  two  neighboring  points 
of  the  curve  (tig.  124).  Draw  the 
tangent  /''/'  and  the  lines  PR  and 
RQ  parallel  to  the  axes,  RQ  and 
FT  intersecting  at  T.  Then,  from 
the  preceding  work, 

PR  =  Ar, 
RQ  =  Ay, 
tan  RPT  =/'(«> 
A'T  =  (tan  RPT)  PR  =f'(x)  &x- 


Fig.  124 


The  quantity  f'(^x)Lx  is  called  the  differential  of  y  and  is 
represented  by  the  symbol  dy.    Accordingly 


dy=f(x)b*\ 


(1) 


142  SLOPES  AND  AREAS 

This  definition  is  true  for  all  forms  of  the  function/  (x)  and  is 
accordingly  true  when  y=f(x)  =  x.  In  this  case/'(V)  =  l,  and 
formula  (1)  gives  dx  =  Ax.  (2) 

Substituting  from  (2)  into  (1),  we  have  the  final  form 

dy=f(x)dx.  (3) 

To  sum  this  up :  The  differential  of  the  independent  variable  is 
equal  to  the  increment  of  the  variable  ;  the  differential  of  the  function 
is  equal  to  the  differential  of  the  independent  variable  multiplied  by 
the  derivative  of  the  function. 

It  is  important  to  notice  the  difference  between  Ay  and  dy. 
The  figure  shows  that,  in  general,  they  are  not  equal,  but  that 
they  become  more  nearly  equal  as  Ax  approaches  zero.  Without 
using  the  figure,  we  may  proceed  thus: 

Since  Lim  ~f-  =f'(x\ 

Az  =  0  AX 

where  Lim  e  =  0  ;  and  hence 

Aa;  =  0 

Ay  =f'(x)  Ax  +  eAx  =  dy  +  eAx. 

Ex.  1.    Let  y  =  x3. 

We  may  increase  x  by  an  increment  Ax  equal  to  dx.    Then 
Ay  =  (x  +  dx)3  -  xz  =  3  xHx  +  3  x  (dxf  +  (dxf. 
On  the"  other  hand,  by  definition, 

dy  =  3  x2 dx. 

It  appears  that  A.y  and  dy  differ  by  the  expression  3  x  (dx)2  +  (dx)3,  which 
is  very  small  compared  with  dx. 

Ex.  2.    If  a  volume  v  of  a  perfect  gas  at  a  constant  temperature  is  under 

the  pressure  p,  then  v  =  - ,  where  k  is  a  constant.    Now  let  the  pressure  be 

P 
increased  by  an  amount  A/>  =  dp.    The  actual  change  in  the  volume  of  the 
gas  is  then  the  increment 

a  ,  _       ^      _  &  —  kdp  kfty  I     1 

The  differential  of  v  is,  however, 


p  +  dp      p         p(p  +  dp)  p2  [1  +  'Jp 


kdp 


AREA 


143 


It  is  to  be  emphasized  that  dx  and  dy  are  finite  quantities, 
subject  to  all  the  laws  governing  such  quantities,  and  are  not 
to  be  thought  of  as  exceedingly  minute.  Consequently  both  sides 
of  (3)  may  be  divided  by  dx,  with  the  result 


/'(*)  = 


_dy 


dx 


Then 


That  is,  the  derivative  is  the  quotient  of  two  differentials. 
This  explains  the  notation  already  chosen  for  the  derivative. 

So,  in  general,  the  limit  of  the  quotient  of  two  increments  is  equal 
to  the  quotient  of  the  corresponding  differentials. 

For  let  y=f(x)     and     z  =  $(x). 

Ay=f'(x)Ax  +  e^Ax, 
Az  =  <\>'(x)  Ax  +  e^Ax, 
dy  =  f'(x)dx, 
dz  =  <f>'(x)dxt 

Az      0'OO  +  e,' 

T.     Ay     T.     /'(*)+«,      f'(x)      dy 

Lnn -r1  =  Lim  .,v  J  — l  =  Jm.y  '  =  -# . 


and 


Whence 


Az 


<l>'(x)  +  €2      $(x)       dz 


78.  Area  under  a  curve.  Let  LK  (fig.  125)  be  a  curve  with 
equation  y  =/(*),  and  let  OE  =  a  and  OB  =  b.  It  is  required 
to  find  the  area  bounded 
by  the  curve  LK,  the  axis 
of  x,  and  the  ordinates  at  E 
and  B. 

For  convenience,  we  as- 
sume in  the  first  place  that 
a  <  b  and  that  f(x)  is  positive 
for  all  values  of  x  between  a 
and  />.  We  will  divide  the 
line   EB  into  n  equal  parts 

by  placing  Ax  = 


J/,  .V,  M3  Mt  M6  Ma  J/7  J/a 
Fig.  125 


and  laying  off  the  lengths  EM^  =  MXM^ 
XB  =  Ax.    (In  fig.  125,  n  =  9.) 


144  SLOPES  AND  AREAS 

Let  OMt=  Xj,  OM2=x2,  •  •  •,  OMn_1=  xn_v  Draw  the  ordinates 
ED=f(a),  Mfi~f(xd,  M2B2=f(x,),  . .  .,  ^.jP,,.^/^.,), 
and  BC.  Draw  also  the  lines  DR^  PXB^  B,B3,  •  •  •,  JjJ_j.fi,,,  parallel 
to  OX    Then 

/(a)  Ax  =  the  area  of  the  rectangle  EDB  M , 

f(xt}Ax  =  the  area  of  the  rectangle  M^B^M^ 

/(x„)  Ax  =  the  area  of  the  rectangle  M2B,B3M3J 

/(*B_i)A*  =  the  area  of  the  rectangle  Mn_xPn_xBnB. 

The  sum 

/(a)  Ax  +/(s1)  Ax  +/(.r2)  Ax  +  .  .  .  +/(x,t_1)  Ax         (1) 

is  then  the  sum  of  the  areas  of  these  rectangles  and  equal  to 
the  area  of  the  polygon  EDBXI^B.2  •  •  •  Bn_1Bl_1BnB.  It  is  evident" 
that  the  limit  of  this  sum  as  n  is  indefinitely  increased  is  the 
area  bounded  by  ED,  EB,  BC,  and  the  arc  EC. 

The  sum  (1)  is  expressed  concisely  by  the  notation 

'S'/C^Ax, 

i  =  0 

where  2  (sigma),  the  Greek  form  of  the  letter  S,  stands  for  the 
word  "  sum,"  and  the  whole  expression  indicates  that  the  sum 
is  to  be  taken  of  all  terms  obtained  from  /(xf )  Ax  by  giving  to  i 
in  succession  the  values  0,  1,  2,  3,  •  •  •,  n  —  1,  where  xQ  —  a. 
The  limit  of  this  sum  is  expressed  by  the  symbol 


r. 


f(x)  dx, 
where  |  is  a  modified  form  of  S. 


/(x)  dx  =  Lim  V/(x.)  Ax  =  the  area  EBCD. 

»  =  <*»   t  =  0 

It  is  evident  that  the  result  is  not  vitiated  if  ED  or  B C  is 
of  length  zero. 


AREA 


145 


Ex.    Required  to  find  the  area  bounded  by  the  curve  y  =  — ,  the  axis  of 
x,  and  the  ordinates  x  =  2  and  x  —  3  (fig.  126). 

(1)   We  may  divide  the  axis  of  x  between  x  =  2  and  x  =  3  in  10  parts, 

3-2 

placing  Ax  =     ,„     =  .1. 


10 

We  make  then  the  following  calculation 


Ax 


^ 

a  =  2, 

/(a)A*  =  .08 

xx  =  2.1, 

/(xJAx  =  .0882 

x2  =  2.2, 

/(x2)Ax  =  .0968 

x3  =  2.3, 
*4  =  2.4, 

/(x3)Ax  =  .1058 
/(x4)Ax  =  .1152 

x5  =  2.5, 

/(x5)Ax  =  .1250 

*6  =  2.(5, 

/(x6)Ax  =  .1352 

a-  =27 
x7 t, 

/(.-•-  )Ax  =  .  1458 

•\ 

x8  =  2.8, 

/(*8)Az  =  .1568 

xfl  =  2.9, 

/(z9)A*  =  .1682 

1.2170 

The  first  approximation  to  the  area  is  therefore  1.217,  which 
example,   the   value    of    the 
sum   (1)   for  n  =  10. 

(2)  As  a  better  approxi- 
mation the  Btudent  may 
compute  the  sum  for  n  =  20 

3  —  2 

.05.    The 


forth 


ind  Ax 

20 

result  is  1.2118. 

(3)   If    we   take 

3-2 

100 

calculation  is  very  tedious. 

The      result,      however,     is 

1.26167.     These    successive 


and    Ax 


=  .01, 


100 
the 


Fig.  126 


determinations    appear    to    be     approaching    a    limit.      By    subsequent 
methods    it    will    be    shown    that    this    limit    is    1/j. 

It  is  obvious  that  the  direct  calculation  of  the  sum  (1)  is  very 
tedious,  if  not  practically  impossible,  if  the  number  of  terms  is 
very  large.  Some  other  method  must  be  found  to  determine  the 
limit  of  the  sum  as  n  increases  indefinitely.  This  method  is  fur- 
nished by  the  discussion  in  the  following  sections. 


146 


SLOPES  AND  AREAS 


79.  Differential  of  area.  Let  any  one  of  the  rectangles  of 
fig.  125  be  redrawn  in  fig.  127  and  relettered,  for  convenience, 
MNRP.  Draw  also  QS  and  complete  the  rectangle  MNQS. 
Let  A  denote  the  variable  area 
EMPD.    Then 

MN=Ax,         RQ  =  Ay, 

MNQP=AA, 

MNRP  =  MP  .  MN=  yAx, 

MNQS  =  NQ.  MN 

~  0/  +  %)  Aft 

But,  from  the  figure, 


that  is, 
whence 


MNRP  <  MNQP  <  MNQS ; 
yAx  <  A  A  <  (y  -\-  Ay)  Ax, 

AA 


i\X 


AA 


dA 


Now  as  Ax  approaches  zero  as  a  limit,  —  -  approaches 

Ax  AA         dx 

y  is  unchanged,  and  y  +  Ay  approaches  y.    Hence  ,  which 

lies  between  y  and  y  +  A?/,  also  approaches  y ;  that  is, 

£-»-/<->  a) 

In  the  differential  notation  we  have 

dA  =  f(x)dx.  (2) 

To  find  the  area  it  is  therefore  necessary  first  to  find  a  func- 
tion whose  derivative  is  f(x)  and  whose  differential  is  f(x)  dx. 

80.  The  integral  of  a  polynomial.  The  process  by  which  a 
function  is  found  from  its  derivative  or  its  differential  is  called 
integration,  and  the  result  of  the  process  is  called  the  integral  of 
the  derivative. 

Integration  is  expressed  by  the  symbol  I  •  thus, 

ff(x)dx  =  F(x),  (1) 


THE  DEFINITE  INTEGRAL  147 

where  F(x)  is  a  function  of  which  the  derivative  is  f(x).  The 
process  may  be  carried  out  in  the  simpler  cases  by  reversing  the 
rules  for  differentiation.    Thus, 

I  2  xdx  =  x2  +  c,         I  3  x\lx  =  x3+  c, 

by  the  formulas  of  §  74. 

In  these  results  c  may  be  any  constant  whatever,  since  —  =  0. 

In  fact,  any  derivative  has  an  infinite  number  of  integrals  dif- 
fering by  a  constant.    The  most  general  form  of  formula  (1)  is 


/ 


f(x)dx  =  F(x)+C,  (2) 


where  F(x)  is  any  particular  function  whose  derivative  is  f(x) 
and  C  is  any  arbitrary  constant,  called  the  constant  of  integration.' 
To  integrate  a  polynomial  we  need  to  know  that  its  integral  is 
the  sum  of  the  integrals  of  its  terms  and  that  the  integral  of 
each  term  is  found  either  by  the  formula 


./' 


ax"  dx  = +  c 

n+1 

or  by  the  formula  I  adx  =  ax  +  c. 

These  are  simply  the  formulas  of  §  74  reversed. 
Ex.    f  (,:•+  :,,-+  7x  +  B)dx  =  ^  +  •'.;'  +  -  •''"  +  3a:  +  C. 

J  \  o  - 

81.  The  definite  integral.  Return  now  to  the  problem  of  area. 
From  §79,  dA=f(x)dx, 

whence,  by  use  of  §  80,  A  =  F(x)  +  C.  (I) 

This  is  the  area  of  the  figure  EMPJ)  (fig.  127),  in  which  the 
line  MP  can  be  drawn  anywhere  between  ED  and  BC.  But  it'  the 
line  MP  coincides  with  ED,  .1=0  and  x=  a.  Substituting  these 
values  in  (1),  we  have  o  =  F(a^  +  C 

whence  C  =  —  F(a). 


148  SLOPES  AND  AREAS 

Formula  (1)  now  becomes 

A  =  F(xy-F(a). 
The  area  A  becomes  the  area  EBCD  when  x  =  b.    Then 
area  EBCD  =  F(K)  -  F(a). 

This  gives  us  our  desired  method  of  evaluating  the  limit  of 
the  sum  (1),  §  78,  and  may  be  expressed  by  the  formula 


i: 


f(x)dx  =  F(V)-F(a).  (2) 

The  limit  of  the  sum  (1),  §  78,  which  is  denoted  by   /  f(x)dx, 

U  a 

is  called  a  definite  integral,  and  the  numbers  a  and  b  are  called 
the  lower  limit  and  the  upper  limit*  respectively  of  the  definite 
integral. 

This  result  gives  the  following  rule  for  evaluating  a  definite 
integral : 

To  find  the  value  of   I  f(x)  dx,   evaluate    J  f(x)  dx,  substitute 

x  =  b  and  x  =  a  successively,  and  subtract  the  latter  result  from 
the  former. 

It  is  to  be  noticed  that  in  evaluating  j  f(x)dx  the  constant 

of  integration  is  to  be  omitted,  since  —  F(ji)  is  that  constant. 
However,  if  the  constant  is  added,  it  disappears  in  the  sub- 
traction, since 

[^(7>)  +  C]  -  \F(a)  +  C]  =  FQ>)  -  F(a). 

In  practice  it  is  convenient  to  express  F(li)  —  F(a)  by  the 
symbol  [FCx)~\ba,  so  that 


X 


f(x)dx=[F(z~)fa. 


Ex.  The  example  of  §  78  may  now  be  completely  solved.  The  required 
areais  r*x2  ,       Tx313     27       8       19      ,  . 

I    5^nTol=T5-15  =  15=1-- 

*  The  student  should  notice  that  the  word  "limit"  is  here  used  in  a  sense 
quite  different  from  that  in  which  it  is  used  when  a  variable  is  said  to  approach 
a  limit  (§  G8). 


THE  DEFINITE  INTEGRAL 


149 


In  the  foregoing  discussion  we  have  assumed  that  f(x)  is 
always  positive  and  that  a  <  b.  These  restrictions  may  be 
removed  as  follows: 

If  f(x)  is  negative  for  all  values  of  x  between  a  and  b,  where 
a  <  b,  the  graphical  representation  is  as  hi  fig.  128.    Here 

f(a) Ax  =  —  the  area  of  the  rectangle  EMJl^I), 

f(x^)  Ax  =  —  the  area  of  the  rectangle  JM^M^BJ^,  etc., 

so  that  f  f(x)  dx  =  -  the  area  EBCD. 

In  case  /(^)  is  sometimes  positive  and  sometimes  negative, 
we  have   a  combination   of  the  foregoing  results,  as   follows: 


I 


Ifa<b,  the  integral 
f(x)  dx    represents 


the  algebraic  sum  of  the 
areas  bounded  by  the 
curve  y  =f(x),  the  axis 
of  x,  and  the  ordinates 
x  =  a  and  x  =  b,  the 
areas  above  the  axis  of 
x  being  positive  and 
those  below  negative. 

If  a  >  b,  Ax  is  negative,  since  Ax  = The  only  change 

necessary  in  the  above  statement,  however,  is  in  the  algebraic 
signs,  the  areas  above  the  axis  of  x  being  now  negative  and 
those  below  positive.  It  is  usual  to  arrange  the  work  so  that 
Ax  shall  be  positive. 

It  is  obvious,  however,  that 

Caf(x~)dx  =  -  ff(x)dx. 

Jb  J  a 

Also,  from  the  areas  involved, 

Cf(x)  dx  =  f  /<V>  dx  +f  f(x)  dx. 


150  SLOPES  AND  AKEAS 

PROBLEMS 

Find  approximately,  by  a  numerical  calculation,  the  slope  of  each 
of  the  following  curves  at  the  point  given  : 

1.  y  =  x2  at  (2,  4). 

2.  y  =  x2  at  (3,  9). 

3.  y  =  v*  at  (1,  1). 

4.  y  =  xs  at  (2,  8). 

Find  from  the  definition,  without  the  use  of  formulas,  the  deriva- 
tives of  the  following  expressions  : 

7.  4a;3.  9.  x5  —  x.  2^ 

8.  5  x2  +  7x-2.  io.  -.  X* 

x2  12.    V.* 

,     Find  by  the  formulas  the  derivatives  of  each  of  the  following 
polynomials : 

13.  4.r3-3x2  +  2x-l.  15.  xs +  7 x'  -6x3  +  7x-3. 

14.  x4  +  7  x2  -  x  +  3.  16.  ^:6-|ic5  +  |a;4  +  22-7  x. 

17.  Prove  that  the  derivative  of  axs  +  bx2  +  ex  +  e  is  the  sum  of 
the  derivatives  of  its  terms. 

18.  By  expanding  and  differentiating  show  that  the  derivative 
of  (4*  +  3)3  is  12(4x  +  3)2. 

19.  By  expanding  and  differentiating  show  that  the  derivative 
of  (x  +  a)'1  is  n(x  +  a)"-1. 

Find  the  values  of  x  for  which  the  following  expressions  are 
respectively  increasing  and  decreasing,  and  draw  their  graphs : 

20.  x2  +  6  x  —  4.  .  23.  x4  —  2x2  +  l. 

21.  x3-3x2  +  7.  24.  2 x-3- 15;c2  +  36 a; -270. 

22.  x4  +  4x-6.  25.  x3-3x2-9x  +  27. 

26.  If  a  stone  is  thrown  up  from  the  surface  of  the  earth  with  a 
velocity  of  100  ft.  per  second,  the  distance  traversed  in  t  seconds  is 
given  by  the  equation  s  =  100 1  —  16 12.  Find  when  the  stone  moves 
up  and  when  down. 


PROBLEMS  151 

27.  A  particle  is  moving  in  a  straight  line  in  such  a  manner  that 
its  distance  x  from  a  fixed  point  A  of  the  straight  line,  at  any  time 
t,  is  given  by  the  equation  x  =  t3  —  9  t'2  4-  24  t  4-  100.  When  will  the 
particle  be  approaching  A  ? 

28.  A  piece  of  wire  of  length  20  in.  is  bent  into  a  rectangle  one 
side  of  which  is  x.  When  will  an  increase  in  x  cause  an  increase  in 
the  area  of  the  rectangle  and  when  will  it  cause  a  decrease  ? 

29.  In  a  given  isosceles  triangle  of  base  20  and  altitude  10  a  rec- 
tangle of  base  x  is  inscribed.  Find  the  effect  upon  the  area  of  the 
rectangle  caused  by  increasing  x. 

30.  A  right  circular  cylinder  with  altitude  2x  is  inscribed  in  a 
sphere  of  radius  a.    Find  when  an  increase  in  the  altitude  of  the 

cylinder  will  cause  an  increase  in  its  volume  and  when  it  will  cause 
a  decrease. 

31.  A  right  circular  cone  of  altitude  x  is  inscribed  in  a  sphere  of 
radius  a.  Find  when  an  increase  in  the  altitude  of  the  cone  will 
cause  an  increase  in  its  volume  ami  when  it  will  cause  a  decrease. 

32.  On  the  line  3a:  +  y  =  G  a  point  /'  is  taken  and  the  sum  s  of 

the  squares  of  its  distances  from  (5,  1 )  and  (7,  3)  computed.    Find 
the  effect  on  s  caused  by  moving  P  on  the  line. 

Find  the  turning  points  of  the  following  curves  and  draw  the 
curves : 

33.  y  =  2  ;>■»  -  9  x8.  35.   y  =  \  x4  -  2  Xs  +  ], 

34.  y  =  2x*  +  3x*  - 1 2x  - 18.  36.  y  =  x*  -  2  a*  +  4. 

37 .  Find  the  equation  of  the  tangent  to  the  curve  y  —  4  x2  +  4  a-  —  3 
at  the  point  the  abscissa  of  which  is  —  1. 

38.  Find  the  equation  of  the  tangent  to  the  curve  y  =  x3  +  4  x1 
at  the  point  the  abscissa  of  which  is  —  3. 

39.  Show  that  the  equation  of  the  tangent  to  the  curve  y  =  ax2  4- 
2  bx  A-  c  at  the  point  (xv  //t)  is  y  =  2  (axi  4- 1>)  x  —  ax?  +  c. 

40.  Show  that  the  equation  of  the  tangent  to  the  curve  y  =  x3  + 
ax  +  b  at  the  point  (xv  y^  is  y  =  (3  x*  4-  a)x  —  2  x3  4-  b. 

41.  Find  the  area  of  the  triangle  included  between  the  coordinate 
axes  and  the  tangent  to  the  curve  y  =  x3  at  the  point  (3,  27). 


152  SLOPES  AND  AREAS 

42.  Determine  the  point  of  intersection  of  the  tangents  to  the 
curve  y  =  x3  —  5  x  4-  7  at  the  points  the  abscissas  of  which  are 
—  2  and  3  respectively. 

43.  Determine  the  point  of  intersection  of  the  tangents  to  the 
curve  y  =  x3  —  3x4-7  at  the  points  the  abscissas  of  which  are 

2  and  0  respectively. 

44.  Find  the  angle  between  the  tangents  to  the  curve  y  =  x2  — 
4  x  4-  1  at  the  points  the  abscissas  of  which  are  1  and  3  respectively. 

45.  Find  the  angle  between  the  tangents  to  the  curve  y  =  x3  — 

3  x2  4-  4  x  —  12  at  the  points  the  abscissas  of  which  are  —  1  and  1 
respectively. 

46.  Find  the  equations  of  the  tangents  to  the  curve  y  =  x3  4-  x2 
that  have  the  slope  8. 

47 .  Find  the  equations  of  the  tangents  to  the  curve  2  x3  4-  4  x2  — 
x  —  y  =  0  that  have  the  slope  \. 

48.  Find  the  points  on  the  curve  y  =  3  x3  —  4  x2  at  which  it  makes 
an  angle  of  45°  with  OX. 

49.  Find  the  points  on  the  curve  y  =  x3  —  x2  4-  2x  4-  3  at  which 
the  tangents  are  parallel  to  the  line  y  =  3  x  —  7. 

50.  How  many  tangents  has  the  curve  y  =  x3  —  2x2  +  x  —  2 
which  are  parallel  to  the  line  7 x  —  4?/4-28  =  0?  Find  their 
equations. 

51.  Find  approximately  the  area  bounded  by  the  straight  line 
y  ='2x4-3,  the  ordinates  x  =  1  and  x  =  2,  and  the  axis  of  x,  by 
considering  the  area  as  the  sum  of  rectangles  the  bases  of  which  are 
.2  in  the  first  approximation  and  .1  in  the  second  approximation. 
Also  find  the  area  exactly  by  elementary  geometry. 

52.  Find  approximately  the  area  between  the  axis  of  x  and  the 
portion  of  the  curve  y  =  x  —  x2  which  is  above  the  axis  of  x,  by 
considering  the  area  as  the  sum  of  rectangles  the  bases  of  which  are 
.2  in  the  first  approximation  and  .1  in  the  second  approximation. 

53.  Find  approximately  the  area  bounded  by  the  curve  y  =  -> 

the  ordinates  x  =  2  and  x  =  3,  and  the  axis  of  x,  by  considering  the 
area  as  the  sum  of  rectangles  the  bases  of  which  are  .2  in  the  first 
approximation  and  .1  in  the  second  approximation. 


PEOBLEMS  153 

54.  Find  the  area  bounded  by  the  curve  y  =  Va-,  the  ordinates 
x  =  1  and  x  =  4,  and  the  axis  of  x,  by  considering  the  area  as  the 
sum  of  rectangles  the  bases  of  which  are  .5  in  the  first  approxima- 
tion and  .2  in  the  second  approximation. 

55.  Find  by  integration  the  area  described  in  Ex.  51. 

56.  Find  by  integration  the  area  described  in  Ex.  52. 

57.  Find  the  area  bounded  by  the  curve  y  =  x3  —  2ar  +  Sx  —  1, 
the  ordinates  x  =  2  and  x  =  4,  and  the  axis  of  x. 

58.  Find  the  area  bounded  by  the  axis  of  x  and  the  portion  of 
the  curve  y  =  9  —  x2  above  the  axis  of  x. 

59.  Find  the  area  between  the  axis  of  x  and  that  part  of  the 
curve  y  =  10  —  11  x  —  6  x2  which  is  above  the  axis  of  x. 

60.  Find  the  area  between  the  axis  of  x  and  that  part  of  the 
curve  y  =  xs  —  3  x2  —  9  x  +  27  which  is  above  the  axis  of  x. 

61.  Find  the  area  bounded  by  the  axis  of  x  and  the  portion  of 
the  curve  y  =  xa  +  3x2  —  4  below  the  axis  of  x. 

62.  Find  each  of  the  two  areas  bounded  by  the  curve  y  =  150  a;  — 
25  x2  —  xs  and  the  axis  of  x. 

63.  Find  the  area  bounded  by  the  axis  of  x,  the  curve  y  =  2x3  + 
3  a;2  -f  2,  and  the  ordinates  through  the  turning  points  of  the  curve. 

64.  Prove  that  the  area  of  a  parabolic  segment  is  two  thirds  of 
the  product  of  its  base  and  altitude. 

65.  Find  the  area  between  the  parabola  y  =  \  x2  and  the  straight 
line  Sx-2y-4:  =  0. 

66.  Find  the  area  of  the  crescent-shaped  figure  between  the 
curves  y  =  x2  +  5  and  y  =  2  x1  +  1. 


CHAPTER  X 

DIFFERENTIATION  OF  ALGEBRAIC  FUNCTIONS 

82.  Theorems  on  derivatives.  In  order  to  extend  the  process 
of  differentiation  to  functions  other  than  polynomials,  we  shall 
need  the  following  theorems: 

1.  The  derivative  of  a  function  plus  a  constant  is  equal  to  the 
derivative  of  the  function. 

Let  u  be  a  function  of  x  which  can  be  differentiated,  let  c  be 

a  constant,  and  place  ..  _  .,  ._■_  „ 

y  —  u  -f-  c. 

Then  if  x  is  increased  by  an  increment  Ax,  u  is  increased  by 
an  increment  Au,  and  c  is  unchanged.  Hence  the  value  of  y 
becomes  u  +  A^  +  c. 

Whence  Ay  =  (u  +  Ah  +  e)  —  (u  +  e)  =  Au. 

Therefore  —  =  —  i 

Ax      Ax 

and,  taking  the  limit  of  each  side  of  this  equation,  we  have 
dy  __  du 

dx      dx 
Ex.  1.   7/=:4a,-3  +  3. 

2.  The  derivative  of  a  constant  times  a  function  is  equal  to  the 
constant  times  the  derivative  of  the  function. 

.Let  u  be  a  function  of  x  which  can  be  differentiated,  let  c  be 

a  constant,  and  place  „,  __  „. 

y  —  ta. 

Give  x  an  increment  Ax,  and  let  Au  and  Ay  be  the  corre- 
sponding increments  of  u  and  y.    Then 

Ay  =  c  (ii  +  A?f )  —  cu  —  c  Au. 


THEOREMS  ON  DERIVATIVES  155 

TT  Ay        Au 

Hence  — z-  =  p— — , 

Ax        Ax 
and,  by  theorem  3,  §  69, 

T  .     Ay         T  .     Au 
Lmi  — —  =  c  Lim  —  • 

Aa;  Ax 

Therefore  -~-  =  c  — -  > 

aa;        aa; 

by  the  definition  of  a  derivative. 
Ex.  2.    #  =  5(x3  +  3x2  +  l). 

lll  =  5  —  (x3  +  3  x2  +  1)  =  5  (3  x2  +  6  x)  =  15  (x2  +  2  x). 

dx  dx 

3.  The  derivative  of  the  sum  of  a  finite  number  of  functions 
is  equal  to  the  sum  of  the  derivatives  of  the  functions. 

Let  u,  v,  and  w  be  three  functions  of  x  which  can  be  differen- 
tiated, and  let  _  u  +  v  +  UK 

Give  x  an  increment  Ax,  and  let  the  corresponding  increments 
of  u,  v,  w,  and  y  be  Au,  Av,  Aw,  and  A//.    Then 

Ay  =  (u  +  Am  +  v  +  Av  +  w  +  Aw)  —  (?*  +  v  +  w) 

=  Au  +  At'  +Aw; 

,  A?/      Am      At'      Aw 

whence  -f-  =  —  +  —  +  -—  • 

Aa;      Aa;      Aa;      Aa; 

Now  let  Aa;  approach  zero.    By  theorem  1,  §  69, 

T  .     Ay     T  .     An,  T  .     Av     T .     Aw 

Lim  — ^  =  Lim  ■ — -  +  Lim \-  Lim ; 

Aa-  Aa-  Aa-  Aa; 

that  is,  by  the  definition  of  a  derivative, 

dy  _  du  ,dv      dw 
dx      dx      dx      dx 

The  proof  is  evidently  applicable  to  any  finite  number  of 
functions. 

Ex.  3.   y  =  x4  -  3  x3  +  2  x2  -  7  x. 

^  =  4  x3  -  9  x2  +  4  x  -  7. 
dx 


156    DIFFERENTIATION  OF  ALGEBRAIC  FUNCTIONS 

4.  Tlie  derivative  of  the  product  of  a  finite  number  of  functions 
is  equal  to  the  sum  of  the  products  obtained  by  multiplying  the 
derivative  of  each  factor  by  all  the  other  factors. 

Let  u  and  v  be  two  functions  of  x  which  can  be  differentiated, 
and  let  y  _  wu# 

Give  x  an  increment  Ax,  and  let  the  corresponding  increments 
of  u,  v,  and  y  be  Aw,  Av,  and  Ay. 

Then  Ay  =  (u  +  Aii)  (v  +  Av)  —  uv 

=  uAv  +  v  Au  +  Au  •  Av 

Av         Av        Au  ,  Aw     . 
and  -rf-  =  it  —  +  ^  t~  +  -t—  •  Av. 

Ax         Ax        Ax      Ax 

If,  now,  Ax  approaches  zero,  we  have,  by  §  69, 

T  .     Ay         T  .     Av  ,      T  .     Am  ,  ,  .     Am    T .      . 
Lim  — -  =  u  Lim  - — \-v  Lim \-  Lim  — -  •  Lim  Av. 

Ax  Ax  Ax  Ax 


But            L 

im  Av  =  0, 

and  therefore 

dy         dv  ,      du 

Again,  let 

y  =  uvw. 

Regarding  uv  as  one  function  and  applying  the  result  already 

obtained,  we  have 

dy          dw         d  (uv) 

-JL—UV—--\-W        ■       J 

dx           dx             dx 

dw  ,            dv         du] 
=  uv  — — f-  w  \u  —  +  v  — 
dx          L    dx        dx] 

dw   ,  C?U    .  $W 

—  UV—-  +  UW—  +  VW—' 

dx  dx  dx 

The  proof  is  clearly  applicable  to  any  finite  number  of  factors, 
Ex.  4.    y  =  (3  x  -  5)  (a:2  +  1)  xs. 

=  (3  x  -  5)(x2  +  1)(3  x2)  +  (3  x  -  5)x3(2x)  +  (x2  +  l)x3(3) 
=  (18  xs  -  25  x2  +  12  x  -  15)x2. 


THEOREMS  ON  DERIVATIVES  157 

5.  The  derivative  of  a  fraction  is  equal  to  the  denominator  times 
the  derivative  of  the  numerator  minus  the  numerator  times  the  deriva- 
tive of  the  denominator,  all  divided  by  the  square  of  the  denominator. 

Let  y  =  - ,  where  u  and  v  are  two  functions  of  x  which  can  be 

v 

differentiated.    Let  Ax,  Am,  Av,  and  Ay  be  as  usual.    Then 
u  +  Au      u      vAu  —  uAv 


±y 


v  +  Av      v         v2+vAv 


Ah        Av 

v~k u  IT 

Ail        Ax         Ax 

and  — -  =  — ; 

Ax        v~+v  Av 

Now  let  Ax  approach  zero.    By  §  69, 

T  .     Au         T  .     Av 
v  Lim u  Lim  — 

_  .     Ay  A.r  Ax 

Lim  — -  = ; ■ 

Ax  v~+  v  Lim  Ay 

du         dv 

dy        dx         dx 


whence 

T2—  1 

Ex.5.  y  =  -a—± 

x*  +  1 


dx 


V 


dy  _  (.r2  +  l)(2x)-(j-2-l)2j-  _       4:r 
dx  (x2  +  l)s  (.<•-  +  l)2 ' 

6.  If  y  is  a  function  of  x,  then  x  is  a  function  of  y,  and  the  " 
derivative  of  x  with  respect  to  y  is  the  reciprocal  of  the  derivative 
of  y  with  respect  to  x. 

Let  Ax  and  Ay  be  corresponding  increments  of  x  and  y.   Then 
Ax  _   1 
Ay~  Aji 

Ax 

whence  Lim  — —  = 


^      Lim^ 
Ax 
dx      1 
that  is,  -—  =  —-. 

dy     dy 

dx 


158    DIFFERENTIATION  OF  ALGEBRAIC  FUNCTIONS 

7.  If y  is  a  function  of  u  and  u  is  a  function  of  x,  then  y  is 
a  function  of  x,  and  the  derivative  of  y  with  respect  to  x  is  equal 
to  the  derivative  of  y  with  respect  to  u  times  the  derivative  of  u 
ivith  respect  to  x. 

An  increment  Ax  determines  an  increment  Au,  and  this  in  turn 
determines  an  increment  Ay.    Then  evidently 

Ay  _  Ay    Au 

Ax      Au    Ax ' 

whence  Lim  — -  =  Lim  — —  •  Lim  — - ; 

Ax  Au  Ax 


dy  _  dy    du 
dx      du    dx 

1 


2  +  3  x2     2  4  +  6  a:2 


that  is, 

Ex.  6.    y  =  m2  +3h  +  1,  where  u  = 

The  same  result  is  obtained  by  substituting  in  the  expression  for  y  the 
value  of  u  in  terms  of  x  and  then  differentiating. 

This  result  has  an  important  application  to  the  differential.  For 
suppose  we  have       y  =/(>)j         u  =  ^  ^  (1) 

By  substitution,  we  obtain 

y=/[<K*)]  =*•(*>  (2) 

and  the  formula  proved  above  gives  us 

**(*)=/'00'.f(*>  (3) 

By  use  of  §  77  we  obtain  from  (1) 

dy  =f'(u)  du,         du  =  </>'(V)  dx,  (4) 

and  from  (2)  we  have    dy  =  F'(x)dx.  (5) 

It  is  important  to  know  that  the  two  values  of  dy  in  (4) 
and  (5)  agree.  In  fact,  by  means  of  (3)  and  the  second  part 
of  (4),  (5)  becomes 

dy=f(ii)y(i^dx=f(,i)du. 

Hence  it  is  not  necessary,  in  applying  §  77  to  find  a  differential, 
to  ask  whether  x  is  an  independent  variable  or  not. 


DERIVATIVE  OF   UN  159 

83.  Derivative  of  un.    If  u  is  any  function  of  x  which  can  be 
differentiated  and  n  is  any  real  constant,  then 

d(un)  m   ,  die 

ax  dx 

To  prove  this  formula  we  shall  distinguish  four  cases: 
1.  When  n  is  a  positive  integer. 

d  (wn)      d  (wn)    du 


dx  du       dx 

i  du 


(by  7,  §  82) 


=  ™-V  (Byl,  §74) 

2.  When  w.  is  a  positive  rational  fraction. 

Let  n  =  —  i  where  p  and  q  are  positive  integers,  and  place 

y  =  u\ 

By  raising  both  sides  of  this  equation  to  the  qt\\  power,  we  have 

yq  =  up. 

Here  we  have  two  functions  of  x  which  are  equal  for  all 
values  of  x.    If  we  give  x  an  increment  A.r,  we  have 

A(y)  =  AO*)> 
A(yO_Ap/p) 

Ax  Ax 

md  therefore  iSfl^Jfl; 

dx  dx 

whence  qyq~l  -^  =  pu1"1  —  * 

1J       dx      r         dx 

since  p  and  q  are  positive  integers.    Substituting  the  value  of  y 
and  dividing,  we  have 

dx      q  dx 

Hence,  in  this  case  also, 

d(un~)  __      B_1  du ^ 


160    DIFFERENTIATION  OF  ALGEBRAIC  FUNCTIONS 

3.  When  n  is  a  negative  rational  number. 
Let  n  =  —  m,  where  m  is  a  positive  number,  and  place 
,      1 

y  =  U    m  =  —' 


um 

d(um) 

Then                            ft-        p 
dx          ulm 

„    i  du 

mum     — 

dx 

u2m 

(by  5,  §  82) 
(by  1  and  2) 

-m-ldu 

=  —  mu         - — 
dx 
Hence,  in  this  case  also, 

d  CM")         .  i  du 
dx                  dx 

4.  When  n  is  an  irrational  number. 

The  formula  is  true  in  this  case  also,  but  the  proof  will  not 
be  given. 

It  appears  that  1,  §  74,  is  true  for  all  real  values  of  n. 

Ex.  1.  y  =  (x3  +  4  x2  -  5  x  +  7)3. 

^  =  3  (x3  +  4  x2  -  5  x  +  7)2  —  (x3  +  4  x2  -  5  x  +  7) 
dx  dx 

=  3  (3  x2  +  8  x  -  5)(x3  +  4  x2  -  5  x  +  7)2. 

3/—  1  2 

Ex.  2.    w  =  V  x2  +  —  =  xs  +  z-8. 
xd 

rfv  2  _i  _  . 
-^  =  -x  3  _  3x~4 
rfx      3 

2         3 
3^x      *4' 
Ex.  3.    r/  =  (x  +  l)Vx2  +  l. 

^=(x  +  l)^2  +  1)%(x2  +  l)i^  +  1) 
dx  dx  dx 

=  (x  +  1)Q|  (x2  +  I)-  *  •  2  x]  +  (x2  +  1)* 

=  x(x±l1  +  (x2  +  1)i 

(x2  +  1)4 

2  x2  +  x  +  1 

VxHTL 


FORMULAS  161 

dy  =  \t     x    \~*  d  I    x    \ 
dx~3\x3  +  V      dx\x3  +  l) 

_  1  lx3  +  1\*  1  -  2  x3 
3\     X     /     (x3  +  l)2 

l-2x3 
3  xf  (x3  +  1)* 

84.   Formulas.  The  formulas  proved  in  the  previous  articles  are 

a) 

(2) 
(3) 

*£»*_,,*+„*»,  (4) 


d(u  +  c) 

du 

dx 

dx 

d  (cii) 
dx 

du 

d(u  +  v) 
dx 

du      dv 
dx      dx 

dx  dx  dx 

,  /u\  du  dv 

\vj  dx  dx 

dx  v'2. 


dx  __L 
dy~dy 

dx 
dy  _  dy    du 
dx      du    dx 

dy 
dy  du 
dx      dx 

du 

Formula  (9)  is  a  combination  of  (7)  and  (8). 


(5) 


te=nu    £'  (6) 


(7) 
(8) 
00 


162    DIFFERENTIATION  OF  ALGEBRAIC  FUNCTIONS 

85.  Higher  derivatives.    If  y  =f(x),  then  -j-  is  in  general  a 

function  of  x  and  may  be  differentiated  with  respect  to  x.    The 
result  is  called  the  second  derivative  of  y  with  respect  to  x 

and  is  indicated  by  the  symbol  -t~(-^}'  which  is  commonly 

abbreviated  into  ——• 
dx1 

Similarly,  the  derivative   of  the   second   derivative   is   called 

the  third  derivative,  and  so  on.    The  successive  derivatives  are 

commonly  indicated  by  the  following  notation : 

y  =f(x),  the  original  function  ; 

-r~  =f(x),  the  first  derivative  ; 
ax 

-z-(-rM  =  -r4    =f"(z),  the  second  derivative; 
dx\dx/      dx2  v  y 

A-(pL\  =  fl=f»(x\    the  third  derivative; 
dx  \azr/      dx 

dny 

-7-7  =/(B)(^)»  the  nt\\  derivative. 

It  is  noted  in  §  9  that /(a)  denotes  the  value  of /(re)  when 
x  =  a.  Similarly,  /'(a),  f"(a),  f'n(a)  are  used  to  denote  the 
values  of  /'(#),  f"(x),  f"'(x)  respectively  when  x  =  a.  It  is 
to  be  emphasized  that  the  differentiation  is  to  be  carried  out 
before  the  substitution  of  the  value  of  x. 

Ex-  lf/(*)=^h4'find/''(°)- 


+  1 


-*+Q.  +  l 


roo 


2z3-6a;2-6x  +  2 


(x*  +  l)3 
Therefore  /"(0)  =  2. 


86.  Differentiation  of  implicit  functions.   Consider  any  equation 
^  the  form  /(*,</)  =0;  (1) 

By  means  of  this  equation,  if  a  value  of  x  is  given,  values  of  y  are 
determined.   Hence  (1)  defines  y  as  a  function  of  x.  When  (1)  is 


IMPLICIT  FUNCTIONS  163 

solved  for  y,  so  that  y  is  expressed  in  terms  of  x,  y  is  an  explicit 
function.    When  (1)  is  not  solved  for  y,  y  is  an  implicit  function. 
For  example, 

3  x*  -  4  xy  +  5  f-  6  x  +  7  y  -  8  =  0, 

which  may  be  written 

5  tf+  (J-  4 *)y  +  (3 x*~  6 x ~  8)  =  °» 
defines  y  as  an  implicit  function  of  x. 
If  the  equation  is  solved  for  y,  giving 

_  7  +  4  x  ±  V209  +  64  x  -  44  x* 

y  =  -     40 ' 

y  is  expressed  as  an  explicit  function  of  x. 

It  is  possible  to  find  ■—  from  (1)  without  solving  (1),  for  we 

have  in  (1)  a  function  of  x  which  is  always  equal  to  zero.  Hence 
its  derivative  is  zero.  The  derivative  may  be  found  by  use  of 
the  formulas  of  the  previous  articles,  as  shown  in  the  following 
examples : 

Ex.  1.    Given  x2  +  y2  =  5. 

Then  d(*+f>  =  0, 

dx 

that  is,  2x  +  2y^  =  0; 

dx 

,  dii  x 

whence  —  = 

dx  y 

The  derivative  may  also  be  found  by  Bolving  the  equation  for  y.    Then 


±  v  5  -  x1, 


f/y               —  x               x 

^       Vs  -  x2       y 

Ex.  2. 

Given 

f- 

■  xy  -r  1  =  0. 

Then 

d(yz)     <i(*y)  =  Qm 

dx             dx 

Hence 

Q     „r/y           di/ 

3  w2^i-  —  x—  —  ?/  =  0, 

J  dx         dx      J 

and 

dy          y 

dx      3  w2  -  x- 

The  second  derivative  may  be  found  by  differentiating  the 
result  thus  obtained. 


164    DIFFERENTIATION  OF  ALGEBRAIC  FUNCTIONS 

Ex.  3.    If  x2  +  «2  =  5,  we  have  found  —  = 

dx  y 

,72  „  ,7    /„\ 

Therefore 


cPy  _ 
dx*    ' 

dx\y) 

= 

dy 
y  —  x  — 
J          dx 

f 

y-xl- 

-) 

>f 

_ 

y°+x2 

Ex.  4.    If  yz  —  xy  —  1  =  0,  we  have  found    - 


dx      3  y2  —  x 
dy        d(Sy2-x) 


Then 


d"y 


^-i)t-'-i 


dx*~                       Qiy*-xy 

(3y*-xy 

(Sy*      x)       V            y(    6/     - 
V    J         JSf-x      »\3y*-x 

-1) 

$if-xf 

(3  »»-*)■ 

87.  Tangents  and  normals.    It  has  been  shown  in  §  76  that 
the  tangent  to  a  curve  y  =/<V)  at  a  point  (xx,  y^)  is 


where 


1^7  )  ^enotes  the  value  of  -j-  at  (x^,  y^). 
\dx/ 1  dx 


The  normal  to  a  curve  at  any  point  is  the  straight  line  perpen- 
dicular to  the  tangent  at  that  point. 

To  find  the  equation  of  the  normal  we  first  find  the  slope 
of  the  tangent  and  then  use  the  method  of  §  32,  3. 


TANGENTS  AND  NORMALS 


105 


Ex.  1.   Find  the  equations  of  the  tangent  and  the  normal  to  the  parabola 
y-  =  3  x  at  the  points  for  which  x  =  3. 

When  x  =  3,  y  =  ±  3. 

By  differentiation  we  have 

-  /'.'/_  o      „„     <h  _  3 


:.'/ 


/./■ 


r/./' 


Therefore  the  slope  of  the  tangent  at  (3,  3)  is  \,  and  the  slope  of  the 
tangent  at  (3,  —  3)  is  —  \. 

Hence  at  (3,  3)  the  equation  of  the  tan- 
gent is 


3  =  \{x  -  3),     or 


y  +  3  =  0, 


and  the  equation  of  the  normal  is 

y  -  3  =  -  2  (a;  -  3),     or     2x  +  y-9  =  0; 
and  at  (3,  —  3)  the  equation  of  the  tangent  is 

y  +  3  =  -  I  {.>■  -  3),     or     a?  +  2y  +  3  =  0, 
and  the  equation  of  the  normal  is 
y  +  3  =  2  (./•  -  3),     or     2  z  —  y  -  9  =  0. 

Ex.  2.  Prove  that  the  norma]  to  a  pa- 
rabola at  any  point  makes  equal  angles 
with  the  axis  of  the  parabola  and  the  focal  radius  drawn  to  the  point. 

Lei  /'(('p  //i)  '"'  an.v  point  of  the  parabola  y*  =  ipx  (fig.  129),  and  let 
F(p,  0)  be  the  focus.  Then  11\  is  the  focal  radius  of  /',.  and  let  /',.Y  be 
the  normal  to  the  parabola.   To  prove  ZI'\I\  —  Z/'/yY. 

By  differentiation  we  have  2  y—  =  1  />,  whence  the  slope  of  the  tan- 

°  p  'Ij  v 

gent  at  Pl  is  —  and  the  Blope  pf  the  normal  is  —  t— •    It  follows  that 

!l\  -  J' 


tan  F2ITPX 

8ji 

Slope  of  FP.  =  -^— ; 
•'i  -  P 

therefore 

1_^L/_^I_\ 
-  Ill  (*1  +  P) 

2p 

if  we  replace  yf  by  4_pz1,  since  yx2  =  ipxv  Px  being  a  point  of  the  parabola. 
Since  tan  FPXJY  =  tan  FNPV  the  angles  are  equal. 


1GG    DIFFEEENTIATION  OF  ALGEBRAIC  FUNCTIONS 


The  angle  of  intersection  of  two  curves  is  the  angle  between 
their  respective  tangents  at  the  point  of  intersection.  The 
method  of  finding  the  angle  of  intersection  is  illustrated  in 
the  following  example: 

Ex.  3.  Find  the  angle  of  intersection  of 
the  circle  x1  +  y2  =  8  and  the  parabola  x2  =  2y. 
The  points  of  intersection  are  Pt  (2,  2)  and 
P2  (-  2,  2)  (fig.  130),  and  from  the  symmetry 
of  the  diagram  it  is  evident  that  the  angles 
of  intersection  at  Px  and  P2  are  the  same. 

Differentiating  the  equation  of  the  circle, 


dy 


dy 


we  have  2  x  +  2  y  —  =  0,  whence  —  = , 

dx  ax  y 

and  differentiating  the  equation  of  the  parab- 

dy 


Fig.   130 


1  and  the  slope 


ola,  we  find  -2.  =  x. 
dx 

Hence  at  P1  the  slope  of  the  tangent  to  the  circle 

of  the  tangent  to  the  parabola  is  2. 

Accordingly,  if  /3  denotes  the  required  angle  of  intersection, 

—  1  —  ° 
tan  ft  =  o    =3, 

or  /3=tan-!3. 

88.  Sign  of  the  second  derivative.    Since  the  second  derivative 

is  the  derivative  of  the  first  derivative,  the  sign  of  — ^  shows 


dy 


dx1 


whether  —  is  an  increasing  or  a  decreasing  function. 

dx  d2v 

The  significance  of  -~  f or  the  graph  y  =/(*)  is  obtained  from 

the  fact  that  -~-  is  equal  to  the  slope ;  hence  — ~  is  the  deriva- 

G/'JO  (X'X 

d2v 
tive  of  the  slope.    Therefore,  by  §  75,  if  — ^  is  positive,  the 


slope  is  increasing ;  if  — ^  is  negative,  the 

slope  is  decreasing.   We  may  have,  accord- 
ingly, the  following  four  cases: 


cW 


1.  f>0, 
dx 


0 
Fig.  131 


Since   both   the    ordinate   and   the    slope    are   increasing,  the 
graph  runs  up  toward  the  right  with  increasing  slope  (fig.  131). 


SIGN  OF  THE  SECOND  DEEIVATIVE 

r 


Id" 


2.  ^>0,     g<0. 
dx  dor 

The  graph  runs  up  toward  the  right 

with  decreasing  slope  (fig.  132). 


3.  ^<0, 
dx 


dx2 


>0. 


132 


The  graph  runs  down  toward  the 
right.  The  slope,  which  is  negative,  is 
increasing  algebraically  and  hence  is 
decreasing  numerically  (iig.  133). 


4.  ^<0, 
dx 


dx* 


<0. 


\p08l- 


The  graph  runs  down  toward  the 
right,  and  the  slope  is  decreasing 
algebraically  (fig.  134). 

The  consideration  of  these  cases  leads 

d2y 
to  the  following  conclusion :  If  — '\ 

tive,  the  graph  is  concave  upward  ;  if  ^-™ 
is  negative,  the  graph  is  concave  downward. 

If  a  curve  changes  from  concavity  in  one  direction  to  con- 
cavity in  the  other  direction  at  any  point,  that  point  is  called  a 

d2y 

point  of  infection.    It  follows  that  at  such  a  point      \,  changes 

sign,  either  by  becoming  zero  or  by  becoming  infinite.    These 
two  cases  are  illustrated  in  the  following 
examples : 

Ex.  1.    Examine  the  curve  y  =  -^(x8  —  G  x2) 
for  points  of  inflection. 

dy_l   2_ 
dx  ~  I  X "     *' 


cPy 

dx2 


x-1 


Fig.  135 


The  curve  (fig.  135)   is   concave   downward 
when  x<2,  is  concave  upward  when  x>2,  and  accordingly  then'   is  a 
point  of    inflection   when   x  —  2.     The    ordinate    of    this    point   is    —  1£. 


108    DIFKEPvKNTIATION  OF  ALGEBRAIC  FUNCTIONS- 


Ex.  2.    Examine  the  curve  y  =  (x  —  2)*  for  points  of  inflection. 
dy  _         1  dhj  _  2 


da: 


3  (.,-2)1 
It  is  evident  that 


dx* 


9  (a: -2)1 


co  if  x  =  2,  and  that 
d*y 


no  finite  value  of  x  makes  — 2.  =  0.    If  a:  <  2, 

^  >  0 ;  and  if  a:  >  2,  f—£  <  0.   Hence  the  point 

dx*  dx"  l  Fig.  136 

for  which  x  —  2  is  a  point  of  inflection,  since 

on  the  left  of  that  point  the  curve  is  concave  upward  and  on  the  right  of 

that  point  it  is  concave  downward  (fig.  130).   The  ordinate  of  this  point  is  0. 

89.  Maxima  and  minima.    lif(x)  changes  from  an  increasing 
function  to  a  decreasing  function  (§  75)  when  x  increases  through 


Fig.  137 


the  value  a  and  /(«)  is  finite,  f(a)  is  called  a  maximum  value  of 
/(.>)  (figs.  137,  138);  and  if  f(x)  changes  from  a  decreasing 


O  x=a 

Fig.  139 


O  x=a 

Fig.  140 


function  to  an  increasing  function  when  x  increases  through  the 
value  a  and  f(a)  is  finite,  /(«)  is  called  a  minimum  value  oif(x) 
(figs.  139, 140). 

Since  the  derivative  of  an  increasing  function  is  positive  and 
the  derivative  of  a  decreasing  function  is  negative,  it  follows 


MAXIMA  AND  MINIMA  169 

that  the  derivative  of  the  function  must  change  sign  at  either 
a  maximum  or  a  minimum  value  and  hence  must  become  either 
zero  or  infinity.    Accordingly  we  have  two  cases : 

I.  If  —  =  0  or  oo  ivhen  x  =  a,  and  ~  >  0  when  x<a,  and-~-<0 

dx  ax  ax 

when  x  >  a,f(a)  is  a  maximum  value  of  y  =f(x). 

II.  If  -~  =  0  or  cc  iv hen  x  =  a,  and  ~~  <  0  when  x  <  a,  and 
,  dx  dx 

all 

-~  >  0  wAew  x  >  a,  f(a)  is  a  minimum  value  of  y  =  f(x). 

If,  however,  — -  changes  sign  by  becoming  infinite  and  at  the 

same  time  y  becomes  infinite  (fig.  33),  the  function  is  discon- 
tinuous and  there  is  no  corresponding  maximum  or  minimum 
value. 

In  order  to  apply  the  above  tests  it  is  necessary  to  factor-1-, 
as  shown  in  the  following  examples: 

Ex.  1.   Find  the  maximum  and  the  minimum  value  of 
/(./)  =  x5 -  5  x*  +  5  x3  +  10  x- -  20  x  +  5. 
We  find  f'(x)  =  5  x*  -  20  x3  +  15  x-  +  20  x  -  21  > 

=  5(x2-l)(x2-4x  +  4) 
=  5(x  +  l)(x-l)(x-2)*. 

The  roots  of /'(x)  =  0  are  —  1, 1,  and  2.  As  x  passes  through  —  1,  /"(./) 
changes  from  +  to  —  .  Hence  x  =  —1  gives  /'(./)  a  maximum  value,  namely  -  1. 
As  x  passes  through  +  1, /"(.')  changes  fnnii  —  to  +■  Hence  x  =  +  l  gives 
f(x)  a  minimum  value,  namely  —  4.  As  x  passes  through  2, ./"(')  does  not 
change  sign.  Hence  x  =  2  gives  f(x)  neither  a  maximum  nor  a  minimum 
value. 

Ex.  2.  A  rectangular  box  is  to  be  formed  by  cutting  a  square  from 
each  corner  of  a  rectangular  piece  of  cardboard  and  bending  the  resulting 
figure.  The  dimensions  of  the  piece  of  cardboard  being  20  by  30  in., 
required  the  largest  box  which  can  be  made. 

Let  x  be  the  side  of  the  square  cut  out.  Then  if  the  cardboard  is  bent 
along  the  dotted  lines  of  fig.  141,  the  dimensions  of  the  box  are  30  —  2x, 
20  —  2  x,  x.    Let  y  be  the  volume  of  the  box.    Then 

y  =  x(20-2x)(30-2x) 
=  600  x  -  100  x2  +  4  x3. 

^  =  600  -  200  x  +  12  x2. 
dx 


170    DIFFERENTIATION  OF  ALGEBRAIC  FUNCTIONS 


Equating 

this 

to  zero,  we 

have 

3  a,-2- 

50  a 

+  150  =  0. 

25 

x  =  — 

±5 
3 

Vf-.8.9or 

12.7 

Hence 

dy  _  ,  0 

ZQ\S~        TO 

dx 


12(j:-3.9)(k-12.7). 


dy 


changes  from  +  to  —  as  x 


X 

X 

30-2X 

Cm, 
ll 

©1 

aa:  Fig.  141 

is  through  3.9.   Hence  x  =  3.9 
gives  the  maximum  value  105(3+  for  the  capacity  of  the  box.    x  =  12.7 
gives  a  minimum  value  of  y,  but  this  has  no  meaning  in  the  problem, 
for  which  x  must  lie  between  0  and  10. 

Ex.  3.  y  =  V(x-l){x-2f  =  (x-  1)*  (x  -  2)1 
dy_  3x-4 

dx      3  V(x-lf(x-2) 

—  —  0  when  x  =  f ,  and  changes  from  +  to  —  as  x  passes  through  J. 


il.r 


</.'/ 


Therefore  x  =  J  gives  a  maximum  value  to  the  function.    —  =  co  when 

,  </.£ 

x  =  lor  2.    When  x  =  1,  —  does  not  change 

re- 
sign.   When  x  =  2,  —  changes  from  —  to 
dx 

+  .    Then  x  =  2  gives  a  minimum  value  of 
the  function.    Its  graph  is  in  fig.  142. 

Referring  to  figs.  137  and  139,  we 

see  that  if  -*-  =  0  at  a  maximum  value 
dx 

of  y,  the  curve  is  concave  downward, 

and    that    if   -^  =  0    at    a   minimum 
dx 

value  of  y,  the  curve  is  concave  upward.    Hence  we  may  have 

the  following  two  cases : 

I.  If  -j-  =  0    and  —~  <  0   ivhen  x  =  a,  f(a)  is  a  maximum 

value  of  y  =f(x). 

dv  d^v 

II.  If  -j-  =  0  and  — ^  >  0  when  x  =  a,  f(a)  is  a  minimum 

value  of  y  =f(x). 


Fig.  142 


MAXIMA  AND  MINIMA 


171 


It  is  evident  that  these  tests  can  be  used  to  advantage  when 

it  may  be  difficult  or  impossible  to  factor  ~-i  and  that  they 

fail  if  —4  also  becomes  zero. 
dzr 

Ex.  4.  Light  travels  from  a  point  A  in  one  medium  to  a  point  B  in 
another,  the  two  media  being  separated  by  a  plane  surface.  If  the  velocity 
in  the  first  medium  is  i\  and  in  the  second 
v2,  recpiired  the  path  in  order  that  the  time 
of  propagation  from  .1  to  B  shall  be  a 
minimum. 

It  is  evident  that  the  path  must  lie  in 
the  plane  through  A  and  B  perpendicular 
to  the  plane  separating  the  two  media, 
and  that  the  path  will  be  a  straight  line 
in  each  medium.  We  have,  then,  fig.  143, 
where  MX  represents  the  intersection  of 
the  plane  of  the  motion  and  the  plane 
separating  the  two  media,  and  ACB  rep- 
resents the  path. 

Let  MA  =  a,  KB  =  b,  MN=  c,  and  MC  =  x.    Then  AC=  Vaa  +  x*  and 
CB  =  V(c  —  x)2  +  b'z.   The  time  of  propagation  from  .1  to  B  is  therefore 


M 


X 


Fig.  143 


whence 


and 


dx 

dx2 


Dl-V  <r  + 


M 

e-xy 

+  I 

2 

V2 

C  - 

-  X 

,V(c- 

.<)- 

+  //- 

+ 

},- 

d-t 
Since  — -  is 

dx- 


v^a*  +  x-y      „a[(e-x)»  +  6«]f 

Iways  positive,  the  time  is  a  minimum  when 

I L=^ =  o. 

^Va2  +  x2       f2V(c  -  xf  +  i2 


(1) 


This  equation  may  be  solved  for  x,  but  it  is  more  instructive  to  proceed 

as  follows : 

x  Mi'        .      , 

=  sin  <p, 


Va2  +  x2 


V(c  -  x)2  +  b°- 

Then  equation  (1)  is 

simf/ 


AC 

CX 
CB 


=  sin  \(/. 


y 


172    DIFFERENTIATION  OF  ALGEBRAIC  FUNCTIONS 

Now  <f>  is  the  angle  made  by  A  C  with  the  normal  at  C  and  is  called  the 
angle  of  incidence,  and  if/  is  the  angle  made  by  CB  with  the  normal  at  C 
and  is  called  the  angle  of  refraction.  Hence  the  time  of  propagation  is  a 
minimum  when  the  sine  of  the  angle  of  incidence  is  to  the  sine  of  the 
angle  of  refraction  as  the  velocity  of  the  light  in  the  first  medium  is  to 
the  velocity  in  the  second  medium.  This  is,  in  fact,  the  law  according  to 
which  light  is  refracted. 

In  practical  problems  the  question  as  to  whether  a  value 
of  x  for  which  the  derivative  is  zero  corresponds  to  a  maxi- 
mum or  a  minimum  can  often  be  determined  by  the  nature  of 
the  problem. 

In  Ex.  2  above,  it  is  evident  that  there  must  be  a  maximum  volume  of 
the  box  and  that  there  can  be  no  minimum  value.    Accordingly,  when  we 

have  found  —  =  0  if  x  —  3.9  or  12.7,  since  12.7  is  unreasonable  in  our 
dx 

problem  we  conclude,  without  further  discussion,  that  x  =  3.9  corresponds 

to  the  maximum  volume. 

90.  Limit  of  ratio  of  arc  to  chord.  The  student  is  familiar 
with  the  determination  of  the  length  of  the  circumference  of  a 
circle  as  the  limit  of  the  length 
of  the  perimeter  of  an  inscribed 
regular  polygon.  So,  in  general, 
if  the  length  of  an  arc  of  any 
curve  is  required,  a  broken  line 
connecting  the  ends  of  the  arc  is 
constructed  by  drawing  a  series  of 
chords  to  the  curve  as  in  fig.  144.        f  ^      144 

Then  the  length  of  the  curve  is 
denned  as  the  limit  of  the  sum  of  the  lengths  of  these  chords 
as  each  approaches  zero  and  as  their  number  therefore  in- 
creases without  limit.  The  manner  in  which  this  limit  is 
obtained  is  a  question  of  the  integral  calculus  and  will  not 
be  taken  up  here. 

We  may  use  the  definition,  however,  to  find  the  limit  of  the 
ratio  of  the  length  of  an  arc  of  any  curve  to  the  length  of  its 
chord  as  the  length  of  the  arc  approaches  zero  as  a  limit ;  that  is, 
as  the  ends  of  the  arc  approach  each  other  along  the  curve. 


LIMIT  OF  RATIO  OF  ARC  TO  CHORD  173 

Accordingly,  let  Px  and  P2  (fig.  145)  be  any  two  points  of  a 
curve,  PXP2  the  chord  joining  them,  and  PXT  and  RT  the  tangents 
to  the  curve  at  those  points  re- 
spectively. We  assume  that  the 
arc  PP2  lies  entirely  on  one  side 
of  the  chord  P1P  and  is  concave 
toward  the  chord.  These  condi- 
tions can  in  general  be  met  by 
taking  the  points  Px  and  P2  near 
enough  together.  Then  it  follows 
from  the  definition  that  l 

P1T+TP2>  arc  PJ>,  >  PXP2 ; 

whence  fiI±S>«55>L 

PXP2  PXP2 

If  TR  is  the  perpendicular  from  T  to  PJ^  and  if  the  angles  P„PXT 
and  PJ\T are  denoted  by  a  and  ft  respectively,  then  J[ T—  l[R  sec  a, 
and  TP  =  EP2  sec  ft  =  (  /;/.]  -  PJi)  sec  ft. 

Therefore  P/P+  TP2  =  2>B  sec  a  +  (/?/*  -  ijfi)  sec  ft 

=  iJ/J  sec  ft  -f /;/.'  (sec  a  -  sec  ft), 
,  PXT  +  TJ^     i?i^  sec  ft  +  iffi (sec  a  -  sec  ft) 

=  sec  ft  +  -^— -  (sec  a  —  sec  ft). 

Now,  as  7,*  and  P2  approach  each  other  along  the  curve,  a  and  ft 
both  approach  zero  as  a  limit,  whence  sec  a  and  sec  ft  approach 

PR 

unity  as  a  limit;  and  since  — —  is  always  less  than  unity,  it  fol- 

P  T  +  TP 
lows  that  the  limit  of  — —  is  unity. 

arc  PP. 

Hence  — ^-J— -  lies  between  unity  and  a  quantity  approaching 

1  .  arc  PR 

unity  as  a  limit,  and  therefore  the  limit  of — -  is  unity ;  that  is, 

the  limit  of  the  ratio  of  an  arc  to  its  chord  as  the  arc  approaches 
zero  as  a  limit  is  unity. 


174    DIFFERENTIATION  OF  ALGEBRAIC  FUNCTIONS 


91.  The  differentials  dx}  dy,  ds.  On  any  given  curve  let  the 
distance  from  some  fixed  initial  point  measured  along  the  curve 
to  any  point  P  be  denoted  by  s,  where  s  is  positive  if  P  lies  in 
one  direction  from  the  initial  point 
and  negative  if  P  lies  in  the  oppo- 
site direction.  The  choice  of  the 
positive  direction  is  purely  arbi- 
trary. We  shall  take  as  the  posi- 
tive direction  of  the  tangent  that 
which  shows  the  positive  direction 
of  the  curve  and  shall  denote  the 
angle  between  the  positive  direc- 
tion of  OX  and  the  positive  direction 
of  the  tangent  by  <£. 

Now  for  a  fixed  curve  and  a  fixed  initial  point  the  position 
of  a  point  P  is  determined  if  s  is  given.  Hence  x  and  y,  the 
coordinates  of  P,  are  functions  of  s  which  in  general  are  con- 
tinuous and  may  be  differentiated.    We  will  now  show  that 


dx 
ds 


cos<£, 


dy  .  , 
■—-  =  sin  <p. 
ds 


Let  arc  PQ  =  As  (fig.  146),  where  P  and  Q  are  so  chosen  that 
As  is  positive.    Then  PR  =  Ax  and  RQ  —  Ay,  and 


Ax 
AT 


Ay 

As 


PR     _  chord  PQ 
axcPQ        arc  PQ 

chord  P  Q  „..,_ 

—r ~-  cos  RPQ, 

arc  PQ 

RQ         chord  PQ 


PR 


chord  PQ 


RQ 


avoPQ        arc  PQ 
chord  PQ 


chord  PQ 


arc  PQ 


sin  RPQ. 


Taking    the    limit,    we    have,    since    Lim 


LimRPQ^cf), 


chord  PQ 
arc  PQ 


dx 

— -  =  cos  d>, 
ds 


dy 
ds 


sin  cf>. 


1   and 


(1) 


RATE  OF  CHANGE  175 

If  the  notation  of  differentials  is  used,  equations  (1)  become 
dx  =  ds  •  cos  (f>,         dy  =  ds  •  sin  <f> ; 

whence,    by    squaring    and    adding,    we    obtain    the    important 

equation  ,     „      ., 

ds  =dx  +dy.  (2) 

This  relation  between  the  differentials  of  x,  y,  and  s  is 
often  represented  by  the  triangle  of  fig.  147.  This  figure  is 
convenient  as  a  device  for  memorizing  formula  (1),  but  it 
should  be  borne  in  mind  that  EQ  is  not 
rigorously  equal  to  dy  (§  77),  nor  is  PQ 
rigorously  equal  to  ds.  In  fact,  RQ  =  Ay 
and  I'(i>=As,  but  if  this  triangle  is 
regarded  as  a  plane  right  triangle,  we 
recall  immediately  the  values  of  sin  <£, 
cos  </>,  and  tan  <p  which  have  been  pre- 
viously proved. 

92.  Rate  of  change.    If  y  =/(.>■),  a  change  of  A./-  units  in  ./■ 

causes  a  change  of  Ay  units  in  //,  and  the  quotient     '    gives  the 

Ax 

ratio  of  these  changes.     If  this  ratio  is  equal  to  m,  Ay  =  mAx\ 

that  is,  the  change  in  y  is  on  times  the  change  in  x.    Hence,  if 

m  were   independent   of  A./-,  a  change   of  one   unit   in  u-  would 

cause  a  change  of  m  units  in  //,  and  — *  would  consequently 

measure  the  change  in  y  per  unit  of  change  in  ./•.  But  m  does 
depend  in  general  upon  A.r,  and  hence  does  not  give  an  unam- 
biguous measure  of  the  relative  changes  in  x  and  y.    To  obtain 

such  a  measure,  it  is  convenient  to  take  the  limit  of  -—  as  Ax 

Ax 

approaches  zero  and  to  call  this  limit  the  rate  of  change  of  y 

with  respect  to  x.    We  have  then 

-y-  =  rate  of  change  of  y  with  respect  to  x. 

Ex.  1.  Coefficient  of  expansion.  Let  a  substance  of  volume  v  be  at  a 
temperature  t.  If  the  temperature  is  increased  by  A^,  the  pressure  remain- 
ing constant,  the  volume  is  increased  by  Ac.  The  change  per  unit  of  vol- 
ume is  then  — ,  and  the  ratio  of  this  change  per  unit  of  volume  to  the 


176    DIFFERENTIATION  OF  ALGEBRAIC  FUNCTIONS 

change  in  the  temperature  is  -  —  •  The  limit  of  this  ratio  is  called  the 
coefficient  of  expansion ;  that  is,  the  coefficient  of  expansion  equals 

In  other  words,  the  coefficient  of  expansion  is  the  rate  of  change  of  a  unit 
of  volume  with  respect  to  the  temperature. 

Ex.  2.  Elasticity.  Let  a  substance  of  volume  v  be  under  a  pressure  p. 
If  the  pressure  is  increased  by  A/>,  the  volume  is  increased  by  —  Ay.    The 

change  in  volume  per  unit  of  volume  is  then The  ratio  of  this 

change  per  unit  of  volume  to  the  change  in  the  pressure  is — ,  and 

v  A/> 
the  limit  of  this  is  called  the  compressibility ;  that  is,  the  compressibility 
is  the  rate  of  change  of  a  unit  of  volume  with  respect  to  the  pressure. 
The  reciprocal  of  the  compressibility  is  called  the  elasticity,  which  is 

therefore  equal  to  —  v  —  • 
dv 

In  many  cases  it  is  convenient  to  take  time  t  as  the  inde- 
pendent variable.  Then  — -  and  —  measure  the  rates  at  which 
1  dt  dt 

x  and  y  respectively  are  changing  with  the  time.  If  both  x  and 
y  can  be  expressed  in  terms  of  t,  these  rates  may  be  found  by 
differentiating ;   but  if  y  is  expressed  in  terms  of  x  and  x  is 

expressed  in  terms  of  t,  -~-  may  be  found  by  the  formula 

dy  _  dy  dx 

dt      dx  dt 

which  is  a  special  case  of  (8),  §  84. 

Ex.  3.  A  stone  thrown  into  still  water  causes  a  series  of  concentric 
ripples.  If  the  radius  of  the  outer  ripple  is  increasing  at  the  rate  of  5  ft. 
a  second,  how  fast  is  the  area  of  the  disturbed  water  increasing  when  the 
outer  ripple  has  a  radius  of  12  ft.? 

Let  x  be  the  radius  of  the  outer  ripple  and  A  the  area  of  the  disturbed  water. 

Then  A  =  ttx2 

.  dA       n       dx 

and  =  lirx  —  • 

dt  dt 

By  hypothesis,  —  =  5. 

dA 
Therefore  — —  =  10  ttx  ; 

dt 

and  when  x  =  12, 

dA 

=  120  7r,  the  required  rate. 

tit 


RECTILINEAR  MOTION  177 

This  problem  may  also  be  solved  by  expressing  J.  directly  in  terms  of  t. 
By  the  conditions  of  the  problem,   x  =  5  t, 


and  therefore 

A  =  25tt12; 

whence 

dt 

When  x  =  12,  t  =  2i  and — 
dt 

=  120  7r,  as  before. 

93.  Rectilinear  motion.  An  important  application  of  the  con- 
cept of  a  derivative  is  found  in  the  definition  of  the  velocity  of 
a  moving  body.  We  shall  confine  ourselves  in  this  article  to 
rectilinear  motion,  that  is,  to  motion  which  takes  place  in  a 
straight  line. 

Let  a  body  move  along  a  straight  line  AL  (fig.  148),  and  let 
its  distance  from  a  fixed  point  A,  at  any  time  t,  be  denoted  by  8. 

Then,  if  the  body  is  at  the  point  P  at  

the  time  t,  AP=s.  *      ?        $  L 

The  velocity  of  the  body  is  then  defined  FlG-  148 

as  the  rate  of  change  of  the  distance  8  with  respect  to  the  time  t. 

More  in  detail,  if  t  is  increased  by  the  increment  At,  let  s 

As 
be  increased  by  the  amount  As=PQ.    Then  —  is  the  average 

velocity  of  the  body  dining  the  period  At.    Since  this  average 

velocity  depends  in  general  upon  the  magnitude  of  At,  we  take 

As 
the  limit  of  —  as  At  approaches  zero,  and  call  this  limit  the 
At 

velocity  of  the  body  at  the  point  P. 

Hence,  if  v  denotes  the  velocity, 

T  .     As      ds 
v  =  Lim  -—  =  —-• 

At      dt 

If  the  velocity  is  constant  and  equal  to  Jc,  the  motion  is  said 
to  be  uniform,  and  s  =  kt. 

We  note  that  if  v  >  0,  an  increase  of  time  corresponds  to  an 
increase  of  s,  while  if  v  <  0,  an  increase  of  time  causes  a  decrease 
of  s.  Consequently,  the  velocity  is  positive  when  the  body  moves 
in  the  direction  in  which  s  is  measured,  and  negative  if  it  moves 
in  the  opposite  direction. 


178    DIFFERENTIATION  OF  ALGEBRAIC  FUNCTIONS 


Ex.  1.  If  a  body  is  thrown  up  from  the  earth  with  an  initial  velocity 
of  100  ft.  per  second,  the  space  traversed,  measured  upward,  is  given  by 
the  equation 

.v  =  100  t  - 16  t\ 


Then 


ds 


100  -  32  t. 


When  t  <  '.)  J.,  v>0;  and  when  £  >3  i,  v  <0.    Hence  the  body  rises  for  3^  sec. 
and  then  falls.   The  highest  point  reached  is  100  (3  \)  -  10  (3 1)2  =  156  \. 


Ex.  2.  A  man  standing  on  a  wharf  20  ft.  above  the  water  pulls  in 
a  rope  attached  to  a  boat  at  the  uniform  rate  of  3  ft.  per  second. 
Required  the  velocity  with  which  the  boat  approaches  the  wharf. 

Let  A  (fig.  149)  be  the  position  of  the  man 
and  C  that  of  the  boat.   Let 


AB  = 

We  wish  t 

h  = 

o  li 

=  20, 

AC  =  s,     and     BC 

Now 
therefore 

x  =  Vs2  -  400  ; 
dx              s          df 

dt       V*2  -  400  (/ 

But,  by  hypothesis,  s  is  decreasing  at  the  rate  of  3  ft.  per  second ;  there- 
fore —  =  —  3,  and  the  required  expression  for  the  velocity  of  the  boat  is 
(Ac  _       -  3  .9 
dt       Vs2  -  400 


To  express  this  in  terms  of  the  time,  we  need  to  know  the  value  of  s 
when  t  =  0.    Suppose  this  to  be  s0 ;  then 


and 


s  =  st 
dx_ 
dt  ~ 


3t 


3  sn  +  0  t 


Vsj  -  400  -  o  s7t  +  y  p 


When  the  motion  of  the  body  moving  in  a  straight  line  is  not 
uniform,  the  velocity  at  the  end  of  an  interval  of  time  is  not 
the  same  as  at  the  beginning.    Then,  if  v  +  Av  denotes  the  velocity 

of  the  body  at  Q  (fig.  148),  —  is  the  average  change  of  velocity 

per  unit  of  time  during  the  period  A*.    The  limit  of  this  ratio  is 
called  the  acceleration;  that  is,  the  acceleration  of  a  body  moving 


RECTILINEAR  MOTION  179 

in  a  straight  line  is  the  rate  of  change  of  the  velocity  with  respect 
to  the  time.    Hence,  if  a  denotes  the  acceleration, 

_dv_<l/<h\_cPs 
a~  dt~  dt\dt)~  dt*' 

If  a  is  constant,  the  motion  is  said  to  be  uniformly  accelerated, 
and  v  =  Jet,  where  k  is  constant. 

When  a  is  positive,  an  increase  of  t  corresponds  to  an 
increase  of  v.  This  happens  when  the  body  moves  with 
increasing  velocity  in  the  direction  in  which  8  is  measured 
or  with  decreasing  velocity  in  the  direction  opposite  to  thai 
in  which  a  is  measured. 

When  a  is  negative,  an  increase  of  t  causes  a  decrease  of  v. 
This  happens  when  the  body  moves  with  decreasing  velocity  in 
the  direction  in  which  8  is  measured  or  with  increasing  velocity 
in  the  direction  opposite  to  that  in  which  8  is  measured. 

The  force  which  acts  on  a  moving  body  is  measured  by  the 
product  of  the  mass  and  the  acceleration.  'Inns,  if  /•'  is  the  force, 
and  hi  the  mass  of  a  body  moving  in  a  straight  line, 

dv         dh 

/'  =  ma  =  m  —  =  m  — -  • 
dt         <it- 

From  this  it  appears  thai  a  force  is  considered  positive  or  nega- 
tive according  as  the  acceleration  it  produces  is  positive  or  nega- 
tive. Hence  a  force  is  positive  when  it  acts  in  the  direction  in 
which  s  is  measured  and  negative  when  it  acts  in  the  opposite 
direction. 

Ex.  3.    Let  s  =A  +  Bt  +  \  Ct2. 

Thou  r  =  Jl  +  t  7, 

a  =  C, 
and  F=mC. 

If  s0  and  i'0  denote  the  values  of  s  and  v  when  t  =  0,  we  have,  from  the 
last  equations, 

so=A>         ro  =  L> 

and  the  original  equation  may  be  written 

S  =  sn  +  vj  +  1  at\ 


180    DIFFERENTIATION  OF  ALGEBRAIC  FUNCTIONS 


94.  Motion  in  a  curve.  When  a  body  moves  in  a  curve,  the 
discussion  of  velocity,  acceleration,  and  force  becomes  more  com- 
plicated as  the  directions  as  well  as  the  magnitudes  of  these  quan- 
tities need  to  be  considered.  We  shall  not  discuss  acceleration 
and  force,  but  will  notice  that  the  definition  for  the  magnitude 
of  the  velocity,  or  the  speed,  is  the 
same  as  before,  namely, 
ds 

— *• 

where  s  is  distance  measured  on  the 
curved  path,  and  that  the  direction 
of  the  velocity  is  that  of  the  tangent 
to  the  curve. 

Also  as  the  body  moves  along  a  curved 
path  through  a  distance  PQ  =  As  (fig.  150),  x  changes  by  an  amount 
PR  =  Ax  and  y  changes  by  an  amount  BQ  =  Ay.   We  have  then 


Ax 


ds 


Lim  —  =  —  =  v  =  velocity  of  the  body  in  its  path, 


T  .     Ax 
Lim——  = 

At 

dx 

dt  ' 

=  vx  =  component  of  velocity  parallel  to  OX, 

Lim  -^  = 

At 

dy 
~dt~ 

=  vy  =  component  of 

velocity  parallel  to  OY. 

Otherwise  expressed,  v  represents  the  velocity 
velocity  of  the  projection  of  P  upon  OX,  and  vy 
of  the  projection  of  P  on  OY. 

Now,  by  (8),  §  84,  and  by  §  91, 

of  P,  vx  the 
the  velocity 

dx      dx 

x      dt      ds 

ds 
di 

=  V  cos  <£, 

(1) 

and 

dy      dy 
y      dt      ds 
=  v  sin  <j). 

ds 
'  di 

(2) 

Squaring 

and  adding,  we  have 

v2  =  vl  -1-  V*. 

(3) 

PROBLEMS  181 

Ex.  If  a  projectile  starts  with  an  initial  velocity  v0  in  an  initial  direction 
which  makes  an  angle  a  with  the  axis  of  x  taken  as  horizontal,  its  position 
at  any  time  t  is  given  by  the  parametric  equations 

x  =  v0t  cos  a,     y  =  v0t  sin  a  —  \  gt2. 

Find  its  velocity  in  its  path. 

We  have  vx  =  —  =  t'0  cos  a, 

at 

(hi 
vy  —  -ft~  v«  sm  a  ~~  S** 

Hence  v  =  Vv2  —  2  yv0t  sin  a  +  <j-t-. 

PROBLEMS 

Find  -^  in  each  of  the  following  cases : 
ax 

1.  y  =  (2a;4-3)(a:2  +  3a;-l). 

2.  y  =  (a;2  +  4  x  -  3)  (3  x*  +  12  a;  +  12). 

x  +  a  IS.  y=(s»-l)« 

3.  y  =  ; ;'  

14.  //  =  ■\/4.»-3  +  (')./•--  5. 

15.  y  =  ^x*  +  x2-2x. 
3 


a; 

—  a. 

a;2 

-4 

x2 

+  4 

X3 

4 

—  a; 

2; 

r2  +  4  r  _  3 

3, 

r2+6. 

B  +  6 

.,■- 

+  *- 

2 

17.   // 


y  =  ^rnr—rk'  .,- +  4  .,•  + 1 

6 


18.  y  = 


2       3  19-  y  =  (3*-l)%r-l)» 

8.  y==3a;*  +  2a;*- 


./•  ■ 


xf  20.  y  =  (l-2a-2)2(x2-3a;4-l). 


9       2  21.  y  =  (x  —  1)-Vx*+T. 

x       a:8  2  x  —  1 

22.  2/  = 


10.  ,=  V,-— .  23.  ,  =  (.2-2,  +  3)t(,3  +  l)l- 

ii.  ysB^?_^S+ *  +  1.  24.  ^ViT^'  +  VT^. 

^      ^  25.   y  = ^=. 

12.  y  =  (2  x3  4-  3  x2  +  6)3.  a;  +  Va;2  -  1 


182    DIFFERENTIATION  OF  ALGEBRAIC  FUNCTIONS 


28.7/-— -7==-  31.   y 


x  +  Va2  +  x2  x  -  Va2  +  x2 

Find  -j-  from  each  of  the  following  equations  : 

32.  :rl  +  2  ,ry  +  if  =  0.  34.   xy  =  (x  +  y)2. 

33.  x5-\-5x4y  —  10xyi  +  y5=0.  35.    (x  +  ?/)*  +  (a;  —  yft  =  a*. 

Find  -—  and  —^  from  each  of  the  following  equations : 

CttAs  CbdU 

36.  3  x2  +  y2  =  1.  39.  sc2  +  x?/  +  y2  =  0. 

37.  a-5  +  y5  =  a5.  40.  y3  =  a  (a;2  +  y2). 

38.  x^  +  ^  =  a?.  41.  xy2  =  a;  +  y. 

42.  Find  the  equations  of  the  tangent  and  the  normal  to  the 
curve  5  ax2  —  4  x2y  =  4  y3  at  the  point  (2  a,  a). 

43.  Find  the  equations  of  the  tangent  and  the  normal  to  the 

■>    ■  -.  "  -  x    ,  L,  .    .    /      3  a    6  a 

strophoid  y  =  ±  x-v  — — —  at  the  point  I  —  —  >  — 

44.  Find  the  point  at  which  the  tangent  to  the  curve  y  =  x8  at 
(1,  1)  intersects  the  curve  again. 

45.  Find  the  equations  of  the  tangent  and  the  normal  to  the 
ellipse  3x2  +  5  y2  =  32  at  a  point  the  abscissa  of  which  is  equal  to 
its  ordinate. 

46.  Find  a  point  at  which  the  tangent  to  the  curve  xy  —  5x2  —  4  =  0 
has  the  slope  1. 

47.  Find  the  length  of  the  portion  of  the  normal  to  the  parabola 
y2  =  8  x  at  (2,  4)  included  between  the  axis  and  the  directrix  of 
the  parabola. 

Find  the  equation  of  the  tangent  to  each  of  the  following  curves 
at  the  point  (a? ,  y^: 

48.  y*  =  xb.  50.   xl  +  y$  =  ai. 

49.  Vx  +  Vy  =  Va.  51.  Xs  -f  y3  -  3  ax//  =  0. 


PROBLEMS  183 

52.  Prove   that   the   equation   of   the   tangent   to   the   parabola 
y2  =  kpx  at  the  point  (x ,  y^  is  y1  y  =  2p(x  +  ax). 

53.  If  the  slope  of  a  tangent  to  the  parabola  y2  =  Apx  is  in,  prove 

•               •                             P 
that  its  equation  is  y  =  mx  -\ 

54.  Prove    that    the    equation    of    the    tangent    to    the    ellipse 

-5  +  jt2  =  1  at  the  point  (xv  y^  is  -~  +  ^r  =  1,  and  that  the  equa- 

a2       //'- 
tion  of  the  tangent  to  the  hyperbola  —  —  '-;  =1  at  the  point  (xv  yS) 

a2  b1 

55.  Prove  that  the  equations  of  the  tangents  with  slope  m  to  the 

X2         I/2  ./'"  if 

ellipse  -^  +  7^  =  1  and  the  hyperbola  —2  —  •—  =  1  are  respectively 


y  =  mx  ±  V (/-//<-  +  //'J  and  y  =  mx  ±  ~v<ri/r  —  b2. 

Draw  each   pair  of   the   following  curves    in   one   diagram   and 

determine  the  angles   at   which   they   intersect: 

56.  x  +  y  —  7  =  0,  a2  -  4  x  -  3  //  4-  1  =  0. 

57.  a2  +  y2  -  10  ./•  4-  14  =  0.    ./■-  +  y*  -  8  y  4-  G  =  0. 

58.  2  y2  -  9  x  =  0,   3  x3  4-  4  //  =  0. 

59.  a2  =  4  ay,    2  r2  4-  2  //J  -  5  03  =  0. 

,-3 

60.  v/2  =  — ,   x-  4-  y*  -  2  ax  =  0. 

z  a  —  x 

61.  ;r2  4-  >/2  =  45,    y2  =  12a. 


62.  a-2  4-  y"  -  12  x  +  16  =  0,    v/2 


a" 


4-3 


63.    if  =  a3,   y2  =  -^ 65.   SB*  =  8  if  -  4  ay,  y 


5 -a  '"  y       a24-4«2 

X* 


64.  a2#  =  4,  y  =  •  66.  a//  =  if.    ,f 


^  +  4  ;/  '  *        2a -a 

67.  a2-3,W,   If-y^- 

68.  y2  =  6(a-3),    4^=(a-3)2(a-l). 

69.  Prove  that  the  parabolas  y*  =  4  ax  +  4  a2  and  y2  =  —  4  &a  +  4  b2 
are  confocal  and  intersect  at  right  angles. 


72.  Find  a  point  on  the  ellipse  —2  +  yj  =  1  sucli  that  the  tangent 


184    DIFFERENTIATION  OF  ALGEBRAIC  FUNCTIONS 

70.  Show  that  for  an  ellipse  the  segments  of  the  normal  between 
the  point  of  the  curve  at  which  the  normal  is  drawn  and  the  axes 
are  in  the  ratio  a2 :  V2. 

x2      t/2 

71.  Find  the  coordinates  of  a  point  on  the  ellipse  ~a  +  T2=l 

such  that   the   tangent    there  is    parallel    to   the  line  joining   the 
positive  extremities  of  the  major  and  the  minor  axes. 

"*"  b2 
there  is  equally  inclined  to  the  two  axes. 

73.  Prove  that  the  portion  of  a  tangent  to  an  hyperbola  included 
by  the  asymptotes  is  bisected  by  the  point  of  tangency. 

74.  If  any  number  of  hyperbolas  have  the  same  transverse  axis, 
show  that  tangents  to  the  hyperbolas  at  points  having  the  same 
abscissa  all  pass  through  the  same  point  on  the  transverse  axis. 

75.  If  a  tangent  to  an  hyperbola  is  intersected  by  the  tangents  at 
the  vertices  in  the  points  Q  and  R,  show  that  the  circle  described  on 
QR  as  a  diameter  passes  through  the  foci. 

76.  Prove  that  the  ordinate  of  the  point  of  intersection  of  two 
tangents  to  a  parabola  is  the  arithmetical  mean  between  the  ordi- 
nates  of  the  points  of  contact  of  the  tangents. 

77.  If,  on  any  parabola,  P,  Q,  and  R  are  three  points  the  ordinates 
of  which  are  in  geometrical  progression,  show  that  the  tangents  at 
P  and  R  meet  on  the  ordinate  of  Q. 

78.  Show  that  the  tangents  at  the  extremities  of  the  chord  of  a 
parabola,  which  is  perpendicular  to  the  axis  of  the  parabola  at  the 
focus,  are  perpendicular  to  each  other. 

79.  Prove  that  the  tangents  described  in  Ex.  78  intersect  on  the 
directrix  of  the  parabola. 

80.  Prove  analytically  that  if  the  normals  at  all  points  of  an 
ellipse  pass  through  the  center,  the  ellipse  is  a  circle. 

81.  Prove  that  any  tangent  to  the  parabola  if  =  Apx  will  meet 
the  directrix  and  the  straight  line  drawn  through  the  focus,  per- 
pendicular to  the  axis  of  the  parabola,  in  two  points  equidistant 
from  the  focus. 

82.  Find  in  terms  of  xx  and  p  the  length  of  the  perpendicular 
from  the  focus  of  the  parabola  if  =  ±px  to  the  tangent  at  any 
point   (xv  yx). 


PROBLEMS  185 

83.  If  from  two  given  points  on  the  axis  of  a  parabola  which 
are  equidistant  from  the  focus  perpendiculars  are  let  fall  on  any 
tangent,  prove  that  the  difference  of  their  squares  is  constant. 

84.  Show  that  the  product  of  the  perpendiculars  from  the  foci 
of  an  ellipse  upon  any  tangent  equals  the  square  of  half  the 
minor  axis. 

85.  Find  the  equation  and  the  length  of  the  perpendicular  from 
the  center  of  the  ellipse  —^  +  j^  =  1  to  any  tangent. 

86.  If  two  concentric  equilateral  hyperbolas  are  described,  the 
axes  of  one  being  the  asymptotes  of  the  other,  show  that  they 
intersect  at  right  angles. 

87.  Prove  that  an  ellipse  and  an  hyperbola  with  the  same  foci 
cut  each  other  at  right  angles. 

88.  Prove  that  the  normal  to  an  ellipse  at  any  point  bisects  tin- 
angle  between  the  focal  radii  drawn  to  the  point. 

89.  Prove  that  the  normal  to  an  hyperbola  at  any  point  makes 
equal  angles  with  the  focal  radii  drawn  to  the  point. 

Determine  the  values  of  x  for  which  the   following  curves  are 
(1)  concave  upward ;  (2)  concave  downward  : 

90.  a  =  4  xs  -  6  x8  +  3.  91.  y  =  **  - 12  x°-  +  2. 

Find  the  points  of  inflection  of  the  following  curves  : 

92.  y  =  2a-3  +  9a-2-2x-5.  96     y=      1       +      *_. 

93.  y  =  3a;4-4x»-6!B»  +  4. 

94.  y  =  (x  +  6  a)(x  —  ay.  J  ^     J 

Find  the  turning  points  and  the  points  of  inflection  of  each  of 
the  following  curves  and  then  draw  the  curve : 

99.   y  =  (x  —  2)2(x  +  2).  102.   y  =  x*  -  4  Xs  +  16. 

100.  y  =  x3  -  3«2  -  9x  -  5.        103.  if  =  x(x2  —  4). 

101.  y  =  x  {x  —  l)3. 


186    DIFFERENTIATION  OF  ALGEBRAIC  FUNCTIONS 

104.  It  is  required  to  fence  off  a  rectangular  piece  of  ground 
to  contain  a  given  area,  one  side  to  be  bounded  by  a  wall  already 
constructed.  If  the  length  of  the  side  parallel  to  the  wall  is  x, 
will  an  increase  in  x  cause  an  increase  or  a  decrease  in  the  total 
amount  of  fencing  ? 

105.  The  hypotenuse  of  a  right  triangle  is  given.  If  one  of  the 
sides  is  x,  find  the  effect  on  the  area  caused  by  increasing  x. 

106.  The  stiffness  of  a  rectangular  beam  varies  as  the  product  of 
the  breadth  and  the  cube  of  the  depth.  From  a  circular  cylindrical 
log  of  radius  a  inches,  a  beam  of  breadth  2  a;  is  cut.  Find  the  effect 
on  the  stiffness  caused  by  increasing  x. 

107.  A  right  cone  is  generated  by  revolving  an  isosceles  triangle 
of  constant  perimeter  about  its  altitude.  If  x  is  the  length  of 
one  of  the  equal  sides  of  the  triangle,  will  an  increase  in  x  cause 
an  increase  or  a  decrease  in  the  volume  of  the  cone  ? 

108.  A  gardener  has  a  certain  length  of  wire  fencing  with  which 
to  fence  three  sides  of  a  rectangular  plot  of  land,  the  fourth  side  be- 
ing made  by  a  wall  already  constructed.  Required  the  dimensions 
of  the  plot  which  contains  the  maximum  area. 

109.  A  rectangular  plot  of  land  to  contain  216  sq.  rd.  is  to  be 
inclosed  by  a  fence  and  divided  mto  two  equal  parts  by  a  fence 
parallel  to  one  of  the  sides.  What^a*^  be  the  dimensions  of  the 
rectangle  that  the  least  amount  of  fencing  may  be  required  ? 

110.  A  gardener  is  to  lay  out  a  flower  bed  in  the  form  of  a  sector 
of  a  circle.  If  he  has  20  ft.  of  wire  with  which  to  inclose  it,  what 
radius  will  he  take  for  the  circle  to  have  his  garden  as  large  as 
possible  ? 

V. 

111.  An  open  tank  with  a  square^ase  and  vertical  sides  is  to 

have  a  capacity  of  4000  cu.  ft.    Find  the  dimensions  so  that  the 
cost  of  lining  it  with  lead  may  be  a  minimum. 

112.  A  rectangular  box  with  a  square  base  and  open  at  the  top  is 
to  be  made  out  of  a  given  amount  of  material.  If  no  allowance  is 
made  for  the  thickness  of  the  material  or  for  waste  in  construction, 
what  are  the  dimensions  of  the  largest  box  that  can  be  made  ? 

113.  Find  a  point  on  the  line  y  =  x  such  that  the  sum  of  the 
squares  of  its  distances  from  the  points  (—  a,  0),  (a,  0),  and  (0,  b) 
shall  be  a  minimum. 


PROBLEMS  187 

114.  A  piece  of  wire  12  ft.  in  length  is  cut  into  six  portions,  two 
of  one  length  and  four  of  another.  Each  of  the  two  former  portions 
is  bent  into  the  form  of  a  square,  and  the  corners  of  the  two  squares 
are  fastened  together  by  the  remaining  portions  of  wire,  so  that  the 
completed  figure  is  a  rectangular  parallelepiped.  Find  the  lengths 
into  which  the  wire  must  be  divided  so  as  to  produce  a  figure  of 
maximum  volume. 

115.  The  strength  of  a  rectangular  beam  varies  as  the  product  of 
its  breadth  and  the  square  of  its  depth.  Find  the  dimensions  of  the 
strongest  rectangular  beam  that  can  be  cut  from  a  circular  cylindri- 
cal log  of  radius  a  inches. 

116.  What  are  the  dimensions  of  the  rectangular  beam  of  great- 
est volume  that  can  be  cut  from  a  log  a  feet  in  diameter  and  b  feet 
long,  assuming  the  log  to  be  a  circular  cylinder  ? 

117.  A  log  in  the  form  of  a  frustum  of  a  cone  is  20  ft.  long,  the 
diameters  of  the  bases  being  2ft.  and  1  ft.  A  beam  with  a  square 
cross  section  is  cut  from  it  so  that  the  axis  of  tin-  beam  coincides 
with  the  axis  of  the  log.  Find  the  beam  of  greatest  volume  that  can 
be  so  cut. 

118.  Find  a  point  on  the  axis  of  x  such  that  the  sum  of  its  dis- 
tances from  the  two  points  (1,  2)  and  (4,  3)  is  a  minimum. 

119.  Find  the  point  on  the  circle  jc9  4-  if  =  <r  such  that  the  sum 
of  the  squares  of  its  distances  from  the  two  points  (2  a,  0)  and  (0,  2  a) 
shall  be  the  least  possible. 

120.  A  water  tank  to  hold  300  cu.  ft.  is  to  be  constructed  in  the 
form  of  a  right  circular  cylinder,  the  base  of  the  cylinder  being 
horizontal.  The  tank  is  open  at  the  top,  and  the  material  used  for 
the  bottom  costs  twice  as  much  per  square  foot  as  that  used  for 
the  lateral  wall.  What  are  the  most  economical  proportions  for  the 
tank  ? 

121.  A  tent  is  to  be  constructed  in  the  form  of  a  regular  quadran- 
gular pyramid.  Find  the  ratio  of  its  height  to  a  side  of  its  base 
when  the  air  space  inside  the  tent  is  as  great  as  possible  for  a  given 
wall  surface. 

122.  An  isosceles  triangle  of  constant  perimeter  is  revolved  about 
its  base  to  form  a  solid  of  revolution.  What  are  the  altitude  and 
the  base  of  the  triangle  when  the  volume  of  the  solid  generated  is 
a  maximum  ? 


188    DIFFERENTIATION  OF  ALGEBRAIC  FUNCTIONS 

123.  Required  the  right  circular  cone  of  greatest  volume  which 
can  be  inscribed  in  a  given  sphere. 

124.  The  total  surface  of  a  regular  triangular  prism  is  to  be  k. 
Find  its  altitude  and  the  side  of  its  base  when  its  volume  is  as 
great  as  possible. 

125.  The  combined  length  and  girth  of  a  postal  parcel  is  60  in. 
Find  the  maximum  volume  :  (1)  when  the  parcel  is  rectangular  with 
square  cross  section ;  (2)  when  it  is  cylindrical. 

126.  A  length  I  of  wire  is  to  be  cut  into  two  portions,  which  are 
to  be  bent  into  the  forms  of  a  circle  and  a  square  respectively.  Show 
that  the  sum  of  the  areas  of  these  figures  will  be  least  when  the 
wire  is  cut  in  the  ratio  ir  :  4. 

127.  A  piece  of  galvanized  iron  b  feet  long  and  a  feet  wide  is  to  be 
bent  into  a  U-shaped  water  drain  b  feet  long.  If  we  assume  that 
the  cross  section  of  the  drain  is  exactly  represented  by  a  rectangle  on 
top  of  a  semicircle,  what  must  be  the  dimensions  of  the  rectangle 
and  the  semicircle  that  the  drain  may  have  the  greatest  capacity : 
(1)  when  the  drain  is  closed  on  top  ?    (2)  when  it  is  open  on  top  ? 

128.  A  circular  filter  paper  10  in.  in  diameter  is  folded  into  a 
right  circular  cone.  Find  the  height  of  the  cone  when  it  has  the 
greatest  volume. 

129.  It  is  required  to  construct  from  two  equal  circular  plates  of 
radius  a  a  buoy  composed  of  two  equal  cones  having  a  common  base. 
Find  the  radius  of  the  base  when  the  volume  is  the  greatest. 

130.  Two  towns  A  and  B  are  situated  respectively  10  mi.  and 
15  mi.  back  from  a  straight  river  from  which  they  are  to  get  their 
water  supply,  both  from  the  same  pumping  station.  At  what  point 
on  the  bank  of  the  river  should  the  station  be  placed  that  the  least 
amount  of  piping  may  be  required,  if  the  nearest  points  of  the  river 
to  A  and  B  respectively  are  20  mi.  apart  ? 

131.  A  man  on  one  side  of  a  river,  the  banks  of  which  are  assumed 
to  be  parallel  straight  lines  2  mi.  apart,  wishes  to  reach  a  point  on 
the  opposite  side  of  the  river  and  10  mi.  further  along  the  bank.  If 
he  can  row  3  mi.  an  hour  and  travel  on  land  5  mi.  an  hour,  find  the 
route  he  should  take  to  make  the  trip  in  the  least  time. 


PROBLEMS  189 

132.  A  power  house  stands  upon  one  side  of  a  river  of  width 
b  miles  and  a  manufacturing  plant  stands  upon  the  opposite  side, 
a  miles  downstream.  Find  the  most  economical  way  to  construct 
the  connecting  cable  if  it  costs  m  dollars  per  mile  on  .land  and 
n  dollars  per  mile  through  water. 

133.  At  a  certain  moment  of  time  a  vessel  is  observed  at  a  point 
A,  sailing  in  the  direction  AB  at  the  rate  of  10  mi.  per  hour,  and 
another  vessel  is  observed  at  C,  sailing  in  the  direction  CA  at  the 
rate  of  20  mi.  per  hour.  The  angle  between  AB  and  AC  is  60°,  and 
AC  is  50  mi.    When  will  the  vessels  be  nearest  to  each  other  ? 

134.  A  vessel  is  sailing  due  north  at  the  rate  of  10  mi.  per  hour. 
Another  vessel,  190  mi.  north  of  the  first,  is  sailing  on  a  course 
S.  G0°  E.  at  the  rate  of  15  mi.  per  hour.  When  will  the  distance 
between  them  be  the  least? 

135.  Find  the  least  ellipse  which  can  be  described  about  a  given 
rectangle,  the  area  of  an  ellipse  with  semiaxes  a  and  b  being  irab. 

136.  Find  the  isosceles  triangle  of  greatest  area  which  can  be 
cut  from  a  semicircular  board,  assuming  that  the  base  of  the  triangle 
is  parallel  to  the  diameter. 

137.  Find  the  isosceles  triangle  of  greatest  area  which  can  be  cut 
from  a  parabolic  segment,  assuming  that  the  vertex  of  the  triangle 
lies  in  the  base  of  the  segment. 

138.  The  number  of  tons  of  coal  consumed  per  hour  by  a  certain 
ship  is  0.3  +  0.001  v3,  where  r  miles  is  the  Bpeed  per  hour.  Find  the 
amount  of  coal  consumed  on  a  voyage  of  1000  miles  and  the  most 
economical  speed  at  which  to  make  the  voyage. 

139.  The  fuel  consumed  by  a  certain  .steamship  in  an  hour  is 
proportional  to  the  cube  of  the  velocity  which  would  be  given  to 
the  steamship  in  still  water.  If  it  is  required  to  steam  a  certain 
distance  against  a  current  flowing  a  miles  an  hour,  find  the  most 
economical  rate. 

140.  The  altitude  of  a  variable  cylinder  is  constantly  equal  to  the 
diameter  of  the  base  of  the  cylinder.  If  when  the  altitude  is  8  ft.  it 
is  increasing  at  the  rate  of  3  ft.  an  hour,  how  fast  is  the  volume 
increasing  at  the  same  instant  ? 

141.  Find  where  the  rate  of  change  of  the  ordinate  of  the  curve 
y  =  x3  —  6  x~  +  3  x  +  5  is  equal  to  the  rate  of  change  of  the  slope 
of  the  tangent. 


190    DIFFERENTIATION  OF  ALGEBRAIC  FUNCTIONS 

142.  The  angle  between  the  straight  lines  AB  and  BC  is  60°,  and 
.1 II  is  28  ft.  long.  A  particle  at  A  begins  to  move  along  AB  toward  B 
at  the  rate  of  4  ft.  per  second,  and  at  the  same  time  a  particle  at  B 
begins  to  move  along  BC  toward  C  at  the  rate  of  8  ft.  per  second.  At 
what  rate  are  the  two  particles  approaching  each  other  after  1  sec.  ? 

143.  A  series  of  right  sections  is  made  in  a  right  circular  cone 
of  which  the  vertical  angle  is  90°.  How  fast  will  the  areas  of  the 
sections  be  increasing  if  the  cutting  plane  recedes  from  the  vertex 
of  the  cone  at  the  rate  of  2  ft.  per  second  ? 

144.  A  roll  of  belt  leather  is  unrolled  on  a  horizontal  surface  at 
the  rate  of  5  ft.  of  length  per  second.  If  the  leather  is  i  in.  thick,  at 
what  rate  is  the  radius  of  the  roll  decreasing  when  it  is  equal  to 

2  ft.,  if  the  roll  is  assumed  to  remain  always  a  true  circle  ? 

145.  A  trough  is  in  the  form  of  a  right  prism  with  its  ends  equi- 
lateral triangles  placed  vertically.  The  length  of  the  trough  is  10  ft. 
It  contains  water  which  leaks  out  at  the  rate  of  1  cu.  ft.  per  minute. 
Find  the  rate,  in  inches  per  second,  at  which  the  level  of  the  water 
is  sinking  in  the  trough  when  the  depth  is  1  ft. 

146.  A  solution  is  being  poured  into  a  conical  filter  at  the  rate  of 

3  cc.  per  second  and  is  running  out  at  the  rate  of  1  cc.  per  second. 
The  radius  of  the  top  of  the  filter  is  10  cm.,  and  the  depth  of  the 
filter  is  30  cm.  Find  the  rate  at  which  the  level  of  the  solution  is 
rising  in  the  filter  when  it  is  one  third  of  the  way  to  the  top. 

147.  A  peg  in  the  form  of  a  right  circular  cone  of  which  the  ver- 
tical angle  is  30°  is  being  driven  into  the  sand  at  the  rate  of  2  in. 
per  second,  the  axis  of  the  cone  being  perpendicular  to  the  surface 
of  the  sand,  which  is  a  plane.  How  fast  is  the  lateral  surface  of  the 
peg  disappearing  in  the  sand  when  the  end  of  the  peg  is  10  in.  below 
the  surface  of  the  sand  ? 

148.  A  body  is  moving  in  a  straight  line  according  to  the  law 
s  =  f  —  9 1-  -f-  15  t.  Find  its  velocity  and  acceleration.  When  is 
the  body  moving  forward  and  when  backward  ? 

149.  A  body  is  moving  in  a  straight  line  according  to  the  law 
s  =  \tf  —  2  t*  +  4 12.  Find  its  velocity  and  acceleration.  When  is  its 
velocity  a  maximum  ?   During  what  interval  is  it  moving  backward  ? 

150.  The  top  of  a  ladder  a  units  long  slides  down  the  side  of  a 
vertical  wall  which  rests  on  horizontal  land.  If  the  velocity  of  the 
top  is  v0,  what  is  the  velocity  of  the  bottom  ? 


PROBLEMS  191 

151.  Two  parallel  straight  wires  are  a  feet  apart.  A  bead  slides 
along  one  of  them  at  the  rate  of  b  feet  per  second.  How  fast  is  the 
bead  approaching  a  fixed  point  on  the  other  wire  ? 

152.  A  boat  with  the  anchor  fast  on  the  bottom  at  a  depth  of 
30  ft.  is  drifting  at  the  rate  of  4  mi.  an  hour,  the  cable  attached  to 
the  anchor  slipping  over  the  end  of  the  boat.  At  what  rate  is  the 
cable  leaving  the  boat  when  50  ft.  of  cable  are  out,  assuming  it 
forms  a  straight  line  from  the  boat  to  the  anchor  ? 

153.  A  lamp  is  60  ft.  above  the  ground.  A  stone  is  let  drop  from 
a  point  on  the  same  level  as  the  lamp  and  20  ft.  away  from  it.  Find 
the  speed  of  the  shadow  on  the  ground  after  1  sec,  assuming  that 
the  distance  traversed  by  a  falling  body  in  tin'  time  t  is  1G  t2. 

154.  A  particle  moves  in  a  plane  so  that  its  coordinates  at  any 

2 
time  t  are  given  by  the   equations  x  =  2 1,  y  =  •    Find  the 

Cartesian  equation  of  its  path,  and  its  velocity  in  its  path. 

155.  Two  points,  having  always  the  same  abscissa,  move  in  such 
a  manner  that  each  generates  one  of  the  curves  y  =  xa  —  12 a?  ■+■  4./- 
and  y  =  x%  —  8 x2  —  8.  When  arc  the  points  moving  with  equal 
speed  in  the  direction  of  the  axis  of  // '.' 

156.  A  particle  is  moving  along  the  curve  y2  =  Ax,  and  when 
x  =  4  its, ordinate  is  increasing  at  the  rate  of  10  ft.  per  second.  At 
what  rate  is  the  abscissa  then  changing,  and  how  fast  is  the  particle 
moving  in  the  curve?  Where  will  the  abscissa  be  changing  ten 
times  as  fast  as  the  ordinate  '.' 

157.  A  ball  is  swung  in  a  circle  at  the  end  of  a  cord  5  ft.  long,  so 
as  to  make  20  revolutions  a  minute.  If  the  cord  breaks,  allowing  the 
ball  to  fly  off  at  a  tangent,  at  what  rate  will  it  be  receding  from  the 
center  of  its  previous  path  T^  sec.  after  the  cord  breaks,  it  no 
allowance  is  made  for  any  new  force  acting? 

158.  The  top  of  a  ladder  32  ft.  long  rests  against  a  vertical 
wall  and  the  foot  is  drawn  along  a  horizontal  plane  at  the  rate  of 
4  ft.  per  second  in  a  straight  line  from  the  Avail.  Find  the  path  of  a 
point  one  fourth  of  the  distanee  from  the  foot  of  the  ladder,  and  its 
velocity  in  its  path  at  any  time  t. 


I  .'/ 


CHAPTER  XI 

DIFFERENTIATION  OF  TRANSCENDENTAL  FUNCTIONS 

95.  Limit  of In  order  to  apply  the  methods  of  the  dif- 

h 
ferential  calculus  to  the  trigonometric  functions,  it  is  necessary 

to  know  the  limit  approached  by  — - —  as  h  approaches  zero  as 

a  limit,  it  being  assumed  that  h  is  expressed  in  circular  measure. 

Let  AOB  (fig.  151)  be  the  angle  A, 
r  the  radius  of  the  arc  AB  described 
from  0  as  a  center,  a  the  length  of  AB, 
p  the  length  of  the  perpendicular  BC  0<f- 
from  B  to  OA,  and  t  the  length  of  the 
tangent  drawn  from  B  to  meet  OA  pro-  ^"""^J/' 

duced  in  D.  .  B' 

Revolve  the  figure  on   OA  as  an  axis 
until  B  takes  the  position  B'.    Then  the  chord  BCB'  =  2p,  the 
arc  BAB'  =  2  a,  and  the  tangent  B'D  =  the  tangent  BD.   Evidently 

BD  +  DB'  >  BAB'  >  BCB', 
whence  t>a>p. 

Dividing  through  by  r,  we  have 
tap 
r      r       r 
that  is,  tan  li  >  h  >  sin  h. 

Dividing  by  sin  h,  we  have 

1  h 

cos  h      sin  h 

,      .  .  _      sin  h      i 

or,  by  inverting,  cos  h  <  — - —  <  1. 

h 


TRIGONOMETRIC  FUNCTIONS  193 


Now  as  h  approaches  zero,  cos  h  approaches  1.    Hence  1 

h 
which  lies  between  cos  h  and  1,  must  also  approach  1 ;  that  is, 


T  .     sin  h     ^ 
Lim  — = —  =  1. 


96.  Differentiation  of  trigonometric  functions.  The  formulas 
for  the  differentiation  of  trigonometric  functions  are  as  follows, 
where  u  represents  any  function  of  x  which  can  be  differentiated  : 

d    .  du 

—  sinw  =  cosw  — ->  (1) 
dx                       dx 

(2) 
(3) 
(4) 

(5) 

d  (hi 

—  esc  u  =  —  esc  u  ctn  u   ■    •  (6 ) 
dx                                     dx 

1.  By  f  8),  §  84,     —  sin  u  =  —  sin  u  •  —  • 
J  w    3  dx  du  -/./■ 

To  find  ■ — ■  sin  u,  we  place  y  =  sin  u. 

(hi 

Then  if  u  receives  an  increment  An,  y  receives  an  increment  Ay, 

wliere  a  •    ,     ,   a    n  o        /     ,  A"\   •    A" 

Ay  =  sin  (?t  +  At< )  —  sin  u  =  2  cos  I  u  -f  — -  1  sin  —  > 

the  last  reduction  being  made  by  the  trigonometric  formula 

.    ,  a  +  b  .    a  —  b 

sm  a  —  sm  6=2  cos  — - —  sin  — - —  • 

Then  we  have 

.    An  .    Au 

sm  -—  sin 

Ay      0        /        Au^  ° 


d 

du 

—-  COS  u  = 

dx 

—  sm  u  — -  * 
dx 

d  L 

o    du 

—  tan  u  = 

sec  w  — - » 
dx 

d    t 

o    rfw 

—  ctn  ?*  = 
dx 

i  —  CSC  U  — -  i 

i 

d 

du 

—  sec  u  = 
dx 

seew  tan  u 

dx 

=  2  COS    w  + 

A**  V         2  /     Ait 


■(-+t)-t 


194  TRANSCENDENTAL  FUNCTIONS 

Let  Aw  approach  zero.    By  2,  §  C>9, 

sin  — — 


Lim  — -  =  Lim  cos  (  u  +  -77-  )  Lim 
Aw 

But  Lim 


Aw  \         2  /  Aw 

by  _dy 


Aw      cZw 

■  Am\ 
Lim  cos  (  w  +  —  1  =  cos  w, 


.    Aw 
sin  — 


and  Lim =  1.  (By  §  95) 


d    . 
Hence  —  sm  w  =  cos  w, 

du 

,  c?    .  c?w 

and  — -  sm  w  =  cos  w  — -  ■ 

dx  dx 


2.  To  find  —  cos  w,  we  write 
dx 


cos  w  =  sin  1  —  —  w 


7T 


Then  —  cos  u  =  —  sin  ( —  —  w ) 

£e       \2        / 


dx  dx 


i-)=S-)  (by(1)) 


/tt        \  (7w 

—"Hi  7*)* 

rfw 
=  —  sin  w  — -  • 
dx 


3.  To  find  — -  tan  w,  we  write 
dx 


sin  u 

tan  w  = 

cos  w 


TRIGONOMETRIC  FUNCTIONS  195 


„,  d  d   sin  u 

Then      —  tan  u  = 

dx  dx  cos  11 


d    .  d 

cos  u  —  sin  u  —  sin  u  —  cos  u 

cosV        X  Cb7(5),§84) 

(cos  w  +  snrit)  — 

COS"  W 


(by  (1)  and  (2)) 


=  sec  u  —  > 
dx 

4.  To  find  — -  ctn  ?/,  we  write 
dx 

GOSU 

ctn  m  =  — 

sin  w 

™,  (7     ,  c?  cosw 

I  hen      —  ctn  u  =  — — : 

dx  dx  sm  u 


d  d    . 

sm  u  —  cos  u  —  cos  ii  ■ —  sm  u 


(by  (5),  §84) 


(by  (1)  and  (2)) 


—  shr  u  —  co&*u  du 

sin*-//  dx 

.,    </// 

=  —  CSC   //  —  • 
dx 

5.  To  find  —  sec  v,  we  write 

dx 

1 
secw  = =  (cosw)    . 

cos  u 
Then      —  sec  u  =  —  (cos  u)~'2  —  cos  u  (by  (6),  §  84) 

=  ™±*a  (by(2)) 

cos2^  ^  ^  y  K  JJ 

<hi 
=  sec  m  tan  u  — -  • 
dx 

6.  To  find  —  esc  ?/,  we  write 

csc  u  —  — —  =  (sin  u)~\ 
sm  u 


196 

TRANSCENDENTAL  FUNCTIONS 

Then 

d                   ,  .      N_2  d    . 
— -  esc  w  =  —  (sm  w)   2  —  sin  w 
ax-                                  -ax- 

(by  (6),  §  84) 

.      du 

=  —  esc  ^  ctn  u  —  • 
dx 

(By  (1)) 

Ex.  1.  y  =  tan  2  a;  —  tan2x  =  tan  2  x  —  (tan  x)2. 

^  =  sec2  2  a:  —  (2  x)  -  2  (tan  a:)  ^-  tan  x 
dx  rfa;  </x 

=  2  sec2  2  x  —  2  tan  x  sec2x. 

Ex.  2.   y  =  (2  sec4x  +  3  sec2x)  sin  x. 


sin  x    8  sec3x  —  (sec  x)  +  6  sec  x  —  (sec  x)    +  (2  sec4x  +  3  sec2x)  —  (sin  x) 

|_  (IX  (IX  J  (-IX 


'III 
dx 

=  sin  x  (8  sec4x  tan  x  +  6  sec2x  tan  x)  +  (2  sec4x  +  3  sec2x)  cos  x 

=  (1  —  cos2x)(8  sec5x  +  6  sec3x)  +  (2  sec3x  +  3  sec  x) 

=  8  sec5x  —  3  sec  x. 

97.  Differentiation  of  inverse  trigonometric  functions.  The 
formulas  for  the  differentiation  of  the  inverse  trigonometric 
functions  are  as  follows: 

1.  —  sin-1w  =  — r^^=  —  when  sin-1w  is  in  the  first  or  the 
dx  Vl-V«k  fourth  quadrant; 

when  sin_1M  is  in  the  second  or 
the  third  quadrant. 

2.  — cos-1m  = _  —  when  cos-1m  is  in  the  first  or  the 

second  quadrant ; 
=  — =  —  when  cos-1?/  is  in  the  third  or  the 
V 1  —  u     x  fourth  quadrant. 

_     d  ,     _j   _      1      du 
dx  1  +  u2  dx 

,     d     ,      ,  1      du 

4.  —-ctn  1u  =  —  - -—-' 

dx  1  +  u  dx 

5.  —  sec-1w  = ^=^  —  when  sec-1?/  is  in  the  first  or  the 

dx  u Vw2 -\dx  third  quadrant . 

dien  sec- hi  is  in  the  second  or 


V«2- 1  dx       the  fourth  quadrant. 


INVERSE  TRIGONOMETRIC  FUNCTIONS  197 

—  csc-1w  = ^=  —  when  csc-1w   is  in  the  first   or 

dx  uy/u*-ldx        the  third  quadrant; 

when  csc-^  is  in  the   second  or 


M    M      -*-  c  X  the  fourth  quadrant. 

The  proofs  of  these  formulas  are  as  follows 


1.  If 

y  =  sin_1w, 

en 

sin  y  =  u. 

Hence,  by  §  96, 

dy     du 

cosy-f-  =  —  , 
ax      ax 

aence 

dy         1     du 
dx      cos  y  dx 

But  cos?/=VT—  u1  wlien  y  is  in  the  first  or  the  fourth  quad- 
rant, and  cos  2/  =  —  Vl  —  u1  when  y  is  in  the  second  or  the  third 
quadrant. 


2.  If 
then 

y  =  C'OS-1W, 

cos  y  —  u. 

Hence 

dy      du 
—  sin  y-f-  =  — - 1 

J  dx      dx 

whence 

vy  = 

and 

dp            1     du 

dx          sin  y  dx 

But  sir 

sin  y  = 

•2   wlu'ii    y   is   in    the   first   or   the    second 

quadrant, 

=  —  Vl  —  u1  when  y  is  in  the  third  or  the 

fourth  quadrant. 

3.  If 

y  =  tan_1w, 

then 

tan  y  =  u. 

Hence 

„    dy      du 

whence 

dy          1      du 
dx      1  +  u'2  dx 

198  TRANSCENDENTAL  FUNCTIONS 


4.  If 

y  =  ctn-1M, 

then 

ctn  y  =  u. 

Hence 

„    du      du 
—  csc2y  -^  —  — » 
dx      ax 

whence 

dy             1      du 
dx         1  +  u2  dx 

5.  If 

y  =  sec_1w, 

then 

sec  y  =  u. 

Hence 

dy     du 
sec2/tan;y-  =  -, 

whence 

dy             1         c 

dx      sec  y  tan  y  c 

till 


But  secy  =  u,  and  tan  ?/ =  Vu2  —  1  when  y  is  in  the  first 
or  the  third  quadrant,  and  tan  y  =  —  vV  —  1  when  y  is  in  the 
second  or  the  fourth  quadrant. 

6.  If  y  =  csc_1M, 

then  esc  y  =  u. 

dy      du 

Hence  —  esc  yctny-f-  =  —  > 

dx      dx 


whence 


dy  _  1         du 

dx  cscy  ctn  y  dx 


But  cse  y  =  u,  and  ctn  y  =  Vw2  —  1  when  y  is  in  the  first 
or  the  third  quadrant,  and  ctn y  =  —  vw2  —  1  when  y  is  in  the 
second  or  the  fourth  quadrant. 

If  the  quadrant  in  which  an  angle  lies  is  not  material  in  a 
problem,  it  will  be  assumed  to  be  in  the  first  quadrant.  This 
applies  particularly  to  formal  exercises  in  differentiation. 


Ex.  1.   y  —  sin-1  Vl  —  .r2,  where  y  is  an  acute  angle. 

^  =  . I ±a-x*)t= — -L=- 

dx      Vl  -  (1  -  x2)    dx  VI  -  x2 

This  result  may  also  be  obtained  by  placing  sin-1  V 1  —  x2  — 


EXPONENTIAL  AND  LOGARITHMIC  FUNCTIONS    199 


Ex.  2.    y  =  sec-1  Vi  x2  +  1  x  +  2. 


dx      V-i  x-  +  4x  +  2  \  (4  r  +  4x-  +  2)-l 
4  x  +  2  1 


( I  ,  -  +  -1 .,-  +  2)  (-2X  +  1)       2  z2  +  2  x  +  1 
i 
98.   Limit  of  (1  +  h)h.    In  obtaining  the  formulas  for  the  differ- 
entiation of  the  exponential  and  the  logarithmic  functions  it  is 

i 
necessary  to  know  the  limit  of  (1  +  //  /'  as  //  approaches  zero,  the 
rigorous  derivation  of  which  requires   methods  which  are  too 
advanced  for  this  book.    We  must  content  ourselves,  therefore, 
with  indicating  somewhat  roughly  the  general  nature  of  the  proof. 

We  begin  by  expanding  (1  4-  h  )''  by  the  binomial  theorem  and 
making  certain  simple  transformations;  thus, 


i,  KH,.  KH(M 


(i+ h)h =i  +  ~  h+ — p— /r+ k '<:!+  •  •  • 

_  1      1      (1-7Q      (t-7>)(1-2/0 


l+[|+[I+""+* 


where  R  represents  the  sum  of  all  terms  involving  h,  Jr.  K\  etc. 
Now  it  may  be  shown  by  advanced  methods  that  as  It  approaches 
zero  R  also  approaches  zero,  and  at  the  same  time 

approaches  e  (§  27).    Hence 

i 

Lim  (1  +  JiY  =  e. 

99.  Differentiation  of  exponential  and  logarithmic  functions. 
The  formulas  for  the  differentiation  of  the  exponential  and  the 
logarithmic  functions  are  as  follows,  where,  as  usual,  u  represents 


200  TRANSCENDENTAL  FUNCTIONS 

any  function  which  can  be  differentiated  with  respect  to  x,  log 
means  the  Napierian  logarithm,  and  a  is  any  constant: 

d  ,  log„£  du  ^^ 

l0goW  =  _6a_  (1) 

dx  u     dx 

d  ,  1  du  a 

-—log  u  =  -—-,  (2) 

dx  u  dx 

d  U  Ul  dU  Q 

—  a  =a  loera  —  ?  (6) 

dx  &     dx  W 

d    v _   u du  . 

dx  dx 

The  proofs  of  these  formulas  are  as  follows : 
1.By(8),§84,|log„«  =  Alog„«.|. 

To  find  —  logaM,  place  y  =  logaM. 

Then  if  u  is  given  an  increment  Aw,  y  receives  an  increment 

Am,  where  A         ,       ,     ,    A    N      , 

A#  =  loga(w  + Am)  — log0M 

Am  .       /.      A?/,\A" 
u  \         u 

the  last  transformation  being  made  by  the  formula p  log  M =  log  l/p. 

Then 

Am      m         \         m  / 

—  may  be  taken 
as  h  of  §  98. 
Hence 

Therefore 
and 


Ay  = 
Am 

u    °  \        u 

3hes  zero,  the  fractioi 

Limfl 

A«  =  0  \ 

AuY" 
+  —      =e. 

M  / 

dy^ 
du 

-  losr„  e 
u     &a 

dy 
dx 

logae  £?M 
M      cfa: 

EXPONENTIAL  AND  LOGARITHMIC  FUNCTIONS    201 

2.  If  y  =  log  u,  the  base   a  of  the  previous    formula  is   e ; 
and  since  logee  =  l,  we  have 


dy 
dx 

_ldu 

u  dx 

3.  If 

y- 

=  aH, 

we  have 

\ogy-. 

=  log  a"  =  u  log  a. 

Hence, 

by 

formula 

(2), 

1  dy 
y  dx 

,         du 

whence 

dy 
dx 

„,         du 

=  «uloga— -• 
dx 

4.  Uy 

=  e 

",  the 

pr 

evious 

dy 
dx 

formula  becomes 
dx 

Ex.  1.    y  =  log(x-  -  1  x  +  5). 

dy  2  x  -  I 


dx      x1  —  4  x  +  5 
Ex.  2.    y  =  e~x2- 

lIl  =  -2xe-^ 
dx 

Ex.  3.    y  —  e-^  cos  far. 

—  =  cos  bx  —  (e~  ar)  +  e-  "*  —  (cos  bx)  =  —  oe"  ^  cos  bx  —  be~  ox  sin  hr 
dx  rfx  '  dx 

=  —  e~  "-r(«  cos  for  +  &  sin  bx). 

100.  Sometimes    the    work   of  differentiating    a   function    is 
simplified  by  first  taking  the  logarithm  of  the  function. 


Ex.  l.   Let 


Vir?- 


l-x2 


Then  iogy  =  log^^— - 

=  Uog(l-.r2)-  ilog(l+.T2). 


1  dy  _ 

X 

a; 

.'/  dx 

1 

-  x"        1  +  X2 

-2x 

(1- 

-aP)(l  + 

x>y 

dy 

-2xy 

dx 

(1- 
(1- 

-x2)(l  + 
-2  a: 

-x2)(l  + 

■'■') 

.^)Vi  + 

x2 

X2 

-2x 

202  TRANSCENDENTAL  FUNCTIONS 

Bence 


and 


(l+,:2)Vl 

This  method  is  especially  useful  for  functions  of  the  form  ?/'', 
where  u  and  v  are  both  functions  of  x.  Such  functions  occur 
rarely  in  practice  and  cannot  be  differentiated  by  any  of  the 
formulas  so  far  given.  By  taking  the  logarithm  of  the  function, 
however,  a  form  is  obtained  which  may  be  differentiated. 

Ex.  2.    Let  y  =  xeiux. 

Then  log  y  =  log(x8in*) 

=  sin  x  ■  log  x. 

Therefore  -  -f-  =  (sin  x)  -  +  cos  x  ■  log  x, 

II  dx  x 

and  —  =  xBiax~1  •  sin  x  +  xBiax  cos  x  •  log  x. 

dx 

101.  Applications.  The  applications  of  differentiation  discussed 
in  the  previous  chapter  are  evidently  applicable  to  problems  in- 
volving transcendental  functions. 

Ex.  1.    Find  the  turning  points  and  the  points  of  inflection  of  the  curve 

(Ex.  5,  §  24)  .         ,    t    •    o 

v  '  s       '  y  =  smi  +  |  sin  2  x. 

We  find  —  =  cos  x  +  cos  2  x  =  2  cos2.c  +  cos  x  —  1  ." 

dx 

=  (2  cos  x  —  1)  (cos  x  +  1). 

Equating  —  to  zero,  we  have  cos  x  =  —  or  cos  x  =  —  1. 

dx  2 

If  cos  ar  =  — ,  x  =  — V  2  »ir  or  —  —  +  2  «7r.   As  x  passes  through  either  of 

these  values,  —  changes  sign,  and  hence  these  values  correspond  to  turning 

dx  o     _ 

points  of  the  curve.  In  fact,  x  =  —  +  2  mr  gives  maximum  values  of  y  —  -  Vs, 

and  x  —  —  ~  +  2?nr  gives  minimum  values  of  y  —  —  '- V3. 


APPLICATIONS 


203 


If  cosx 


dy 


x  =  it  ±  2  nw.    As  x  passes  through  these  values,  —  does 

dx 


not  change  sign.    Hence  these  values  do  not  correspond  to  turning  points 
of  the  curve. 

To  examine  for  points  of  inflection,  we  find 

dx2 


—  sin  x  —  i  sin 


—  sin  x  (4  cos  x  +  1). 


Then 


velocity 
acceleration 


bk  cos  bt, 
--  -  Irk  sin  bt  = 


b% 


This  is  zero  when  x  =  0  +  2nirOTTr  +  2  rnr,  or  when  x  =  cos_1(—  \). 

As  x  passes  through  any  of  these  values,  — {  changes  sign.   These  values 

dx2 
correspond,  therefore,  to  j'oints  of  inflection. 

Ex.  2.    A  particle  of  mass  m  moves  in  a  straight  line  so  that 

.s-  =  /sin  ///, 

where  t  =  time,  s  =  space,  and  b  and  k  are  constants. 

(h_ 
dt  ~ 

df- 
force  =  F=  ma  -  —  mb-s. 

Let  0  be  the  position  of  the  particle  when  I  =  0,  and  let  <~>.\  =  k  and 

OB  —  —  k.    Then  it  appears  from  the  for las  for  .-.and  v  thai  the  particle 

oscillates  forward  and  backward  hetween  Band   I.    It  describes  the  distance 

(>.  I  in  the  time  —  and  moves  from  A'  to  .1  and  bach  to  /<'  in  the  time  - — 
2  b  b 

The  formula  F=—  ndi-s  shows  that  the  parti.de  is  acted  on  l>v  a  force 
directed  toward  0  and  proportional  to  the  distance  of  the  particle  from  0. 

The  motion  of  the  particle  is  called  simple  harmonic  motion. 

Ex.  3.  A  wall  is  to  be  braced  by  means  of  a  beam  which  must  pass  over 
a  lower  wall  b  units  high  and  standing  n  units  in  front  of  the  first  wall. 
Required  the  shortest  beam  which  can  lie  used. 

Let  A B  =  1  (fig.  152)  be  the  beam,  and  let  ('  lie  the  top  of  the  lower 
wall.    Draw  the  line  CD  parallel  to  OB,  and  let  EBC  =  d. 


Then 


l  =  BC+  CA 

=  ECcsc6  +  nr^cO 

=  b  esc  6  +  a  sec  0. 
-  =  —  b  csc  6  ctn  0  +  a  sec  $  tan  i 
a  sin30  —  bcosz0 


Placing 
we  have 


dl 

dd 
a  sin30 


sin2#  cos-0 
=  0,  to  find  the  minimum, 

=  b  cos3  0,  whence  tan  8  =  —  • 


204  TRANSCENDENTAL  FUNCTIONS 

When  6  has  a  smaller  value  than  this,  a  siu36  <  b  cos3  9 ;  and  when  0  has  a 

larger  value,  a  sin3 9  >  b  cos3 6.   Hence  I  is  a  minimum  when  tan#  =  — . 

a? 
Then  /  =  Acac0  +  «sec0 


_  b  \ai  +  bi      a  V  n*  +  6$ 
6t                      a^ 

=  (at  +  6f)t. 

102. 

The  derivatives  in   parametric  representation. 

When  a 

curve 

is  defined  by  the  equations 

we  have,  by  (9),  §  84,          !^  =  £. 
dx      dx 

(1) 

*, 

If  it 

;  is  required  to  find  — ^,  we  may  proceed  as  follows: 

c?2y       c?  (dy\      dt\dx) 
dx2      dx\dx)         dx 

(2) 

Ex. 

di 
For  the  cycloid 

x  =  a(<f>  —  sin  d>), 

y  =  a(l—  cos  <£), 

flty      </d>           a  sin  <f>                  <4 
<7  _    r  _                v       =  ctn     • 

dx       dx       a  (1  —  cos  <f>)              2 

dx2      d<j>\        2) dx 

cosec**  — 

2             2 

a(l  —  cosd>) 

1 

4  a  sin4  ^ 
2 


DIRECTION  OF  A  CURVE 
Formula  (2)  may  be  expanded  as  follows: 


Therefore 


±(dy\ 
dt\dx) 


(dy\      d2y  dx     d2x  dy 


dx 

\dt 


dt2  dt      df  dt 

l_x 

it 


d2y 

d2y 
dt2 

dx      d2x 

~dt  ~~  dt2 

dy 
dt 

dx2 

(IT 

205 


(3) 


103.  Direction  of  a  curve  in  polar  coordinates.  The  direction 
of  a  curve  expressed  in  polar  coordinates  is  usually  determined 
by  means  of  the  angle  between  the 
tangent  and  the  radius  vector.  Let  P 
(r,  0)  (fig.  153)  be  any  point  on  the 
curve,  PT  the  tangent  at  P,  and  -v/r 
the  angle  made  by  PT  and  the  radius 
vector  OP.  Give  d  an  increment  Ad  = 
POQ,  expressed  in  circular  measure, 
thus  fixing  a  second  point  Q(r  +  A>; 
6  +  A6)  of  the  curve. 

To  determine  Ar  describe  an  arc  of 
a  circle  with  center  at  0  and  radius  OP,  intersecting  OQ  at  B. 


Then 
and 


OQ=r+Ar 
BQ=Ar. 


Draw  also  the  chord  PQ  and  the  straight  line  PS  perpendicular 
to  OQ  and  meeting  it  in  S. 

Then  SP=r  sin  Ad, 

OS  =r  cos  Ad, 

SQ=OQ-OS=  r  +  Ar-rcosAd 

Ad 
=  Ar  +  2  r  sin2  —  • 


206  TK A NSC UN  DENTAL   FUNCTIONS 

As  A0  approaches  zero  the  chord  PQ  approaches  the  limiting 
position  PT  and  the  angle  11 QP  approaches  i/r.  But  in  the 
triangle  SPQ, 

SP  rsmA6 

tan  SQP  =  —  = 

SQ      A     ,  „      .  ,A6 

Ar  +  2  r  sin2  — 

sin  Ad 

r 

Ad 


.    Ad 

Ar  .    A0SmT 

+  r  sm 


A6>  '  2      A(9 

2 
Hence,  taking  the  limit,  we  have 

tan^  =  ^-.  (1) 

dd 

If  it  is  desired  to  find  the  angle  MNP  =  <j>,  it  may  he  done  by 

the  evident  relation  ,        ,    '  A  ,~N 

<f>  =  yfr  +  0.  (2) 

104.  Derivatives  with  respect  to  the  arc  in  polar  coordinates. 
In  the  triangle  PQS  (fig.  153), 

SP 


siii  SQP  = 


chord  PQ 
SP  arc  PQ 


mcPQ    chord  PQ 
rsinA0       axcPQ 
As         chord  PQ 
sin  Ad    A6      arcP<? 


Ad       As    chord  PQ 
As  Ad  approaches  zero  SQP  approaches  yjr,  Lim  '  =  1,  and 

chord  PQ        V5      y 

Hence  sin  -v/r  =  r  — -  .  (1) 

as 


CURVATURE 


!07 


By  dividing  (1),  just  obtained,  by  (1)  of  the  previous  article, 

C0S+  =  fs'  (2) 

From  (1)  and  (2)  we  obtain 

rdO  =  sin  -v/r  ds,     dr  =  cos  yjr  ds ; 

whence,  by  squaring  and  adding,  we  obtain 

d$2=dr2+r*de\  (3) 

The  formulas  of  this  and  the  foregoing  article  are  correctly 
represented  by  the  triangle  of  fig.  154,  which  is  a  convenient 
device  for  remembering  the  formulas.  Here 
the  lines  marked  as  differentials  are  really 
increments,  but  as  the  size  of  the  figure  is  re- 
duced, they  become  more  nearly  differentials. 
The  correct  formulas  arc  obtained  by  using 
the  triangle  as  a  straight-line  figure.   We  have 

.        ,       rdd 
tan  yfr 


d*  =  -Jd/+r*dd\ 


cos  yfr 


dr 


sin  yjr  = 


dr 

rd0 
ds  ' 


105.  Curvature.  If  a  point  describes  a  curve,  the  change 
of  direction  of  its  motion  may  be  measured  by  the  change  of 
the  angle  <£  (§91). 

For  example,  in  the 
curve  of  tig.  loo,  if  AJ%  —  s 
and  im  =  As,  and  if  <f>l  and 
</>„  are  the  values  of  <f>  for 
the  points  If  and  i^  respec- 
tively, then  </>.,  —  <^)  j  is  the 
total  change  of  direction 
of  the  curve  between  J^ 
and  i?.  If  03-01  =  A$, 
expressed  in  circular  meas- 
A<£ 

unit   of    the    arc 


Fig.   155 


ure,  the  ratio  — -  is  the  average  change  of  direction  per  linear 


Regarding   </>    as    a  function    of   8   and 


208 


TRANSCENDENTAL  FUNCTIONS 


taking  the  limit  of 

have  -j-i  which 
as 


Acb 


as  As  approaches  zero   as  a  limit,  we 


is  called  the   curvature  of  the   curve   at   the 


and  the  circle  is  a  curve  of  constant 


point  ij".  Hence  the  curvature  of  a  curve  is  the  rate  of  change 
of  the  direction  of  the  curve  with  respect  to  the  length  of  the 
arc   (§92). 

If  ~-  is  constant,  the  curva- 
ds 

ture  is  constant  or  uniform ; 
otherwise  the  curvature  is  va- 
riable. Applying  this  defini- 
tion to  the  circle  of  fig.  156,  of 
which  the  center  is  C  and  the 
radius  is  a,  we  have  A<b  =  PXCPV 
and  hence  As  =  a  A(f>.   Therefore 

Ad>      1      rT  deb      1 

—-£  =  -.    Hence  -?-  =  - 

As      a  ds      a 

curvature  equal  to  the  reciprocal  of  its  radius. 

106.  Radius  of  curvature.  The  reciprocal  of  the  curvature  is 
called  the  radius  of  curvature  and  will  be  denoted  by  p.  Through 
every  point  of  a  curve  we  may  pass  a  circle  with  its  radius 
equal  to  p,  which  shall  have  the  same  tangent  as  the  curve  at 
the  point  and  shall  lie  on  the  same  side  of  the  tangent.  Since 
the  curvature  of  a  circle  is  uniform  and  equal  to  the  reciprocal 
of  its  radius,  the  curvatures  of  the  curve  and  the  circle  are  the 
same,  and  the  circle  shows  the  curvature  of  the  curve  in  a 
manner  similar  to  that  in  which  the  tangent  shows  the  direction 
of  the  curve.    The  circle  is  called  the  circle  of  curvature. 

From  the  definition  of  curvature  it  follows  that . 
_  ds 
P~dj>' 

If  the  equation  of  the  curve  is  in  rectangular  coordinates, 
ds_ 

by  (9),  §84,  p  =  ±. 


dx 


RADIUS  OF  CURVATURE  209 

To  transform  this  expression  further,  we  note  that 
ds  —  dx  +  dy  ; 
whence,  dividing  by  dx    and  taking  the  square  root,  we  have 

Since  ^  =  tan-M^j,  (by  §  91) 

dry 
d<\>  dx2 


dx    1+/^y 


Substituting,  we  have 


\dx) 

HS! 


d2y 

d.r1 


In  the  above  expression  for  p  there  is  an  apparent  ambiguity  of 
sign,  on  account  of  the  radical  sign.  If  only  the  numerical  value 
of  p  is  required,  a  negative  sign  may  be  disregarded. 

Ex.    Find  the  radius  of  curvature  of  the  ellipse  —  +  f-  =  1. 

a-      Ul 


Here 

dy         h-.r 

and 

cPy  _       b* 
dx?          ay 

Therefore 

_  OV+  b*x2)% 
9  "          a*b* 

Another  formula  for  p, 

that  is, 

,  MS1] 

<>>r 

may  be  found  by  defining  q>  as  the  angle  between  0  Y  and  the 
tangent  and  interchanging  x  and  y  in  the  above  derivation. 


210 


TRANSCENDENTAL  EUNCTIONS 


107.  Radius  of  curvature  in  parametric  representation.  If  x 
and  y  are  expressed  in  terms  of  any  parameter  t,  the  radius  of 
curvature  may  be  found  as  follows: 


But 


and 


whence 


ds 
ds  dt 
d4>  =  d$ 

dt 


(By  (9),  §  84) 


ds 

dt~ 

W(iH!)' 

dy 

dy                 dt 

(by  (2),  §  91) 

<t>  = 

=  tan     —  =  tan     — ; 
dx                dx 

"dt 
(dx\  d2y     (dy\  d2x 

d(f> 

\ 

dt)  dt2      \dt)dt2 

dt 

HI 

.     \dt!  . 

\dt) 

dx    d2y      dy    d2x 

dt     dt2       dt     dt2 

dxV    (dyV 
dt       \dt 


Therefore,  by  substitution, 


2-if 


dx    d2y      dy    d2x 
It  '  ~df  ~  ~di  '  df 


Ex.    Find  the  radius  of  curvature  of  the  cycloid 

x  =  a<f>  —  a  sin<£, 

II  =  a  —  a  cos  <jf>. 
Here  the  parameter  t  of  the  general  formula  is  replaced  by  cj>. 


Therefore 


RADIUS  OF  CURVATURE 
dx 


211 


tty 


a  —  a  cos  <f>, 


d2x  .      , 
=  a  sin  g>  : 


a  sin  (p, 


dhi  , 

—  =  «cos<^. 


Hence,  by  substitution,    p  — 


[q2  (1  —  cos  <fr)2  +  rr  sin2  <ft]  * 

«(1  —  cos<£)  •  «cos<£  —  f/  sin  $(//  sin  <£) 
22  a  (1-  cosc£)' 


=  2*a-(2siii2|) 


=  4  a  sin 


108.  Radius  of  curvature  in  polar  coordinates.    For  a  curve 
expressed  in  polar  coordinates  the  radius  of  curvature  may  be 

found  as  follows: 


(h_dd 

d<f>~  d$ 

dd 


(By  (9),  §84) 


From  §  104, 

i-^+S) 

and,  from  §  103, 

4>  =  yfr  +  6. 

Then 

tf<9             <Z0             t*0 

r 

.,.©'-• 

d2r 
d0* 

«$ 


Substituting  these  values  and   simplifying,  we  have  as  the 
required  formula,  3 

Naff 


P  = 


r  + 


d& 


212  TRANSCENDENTAL  FUNCTIONS 

Ex.  Find  the  radius  of  curvature  of  the  cardioid  r  =  a(l  —  cos  $). 

Here  — -r  =  a  sin  6  aud  —  =  a  cos  0. 

(10  do'1 


Therefore       o  = 


[q2(l  -  cos  6)"  +  <r  sin2  flp 


«2(1  -  cos  Bf  +2  a2  sin2  6  -  a  (1  -  cos  5)  a  cos  0 

_[2«2(l-cosfl)]f  =  2*a  .£ 

«2(3-3cos0)     ~    3    (1      COS^}  ' 

P  =  1(2  ar)i 

PROBLEMS 

Find  y-  in  each  of  the  following  cases  : 

1.  y  =  isin42a\  2.  y  =  i  sin5  3  a;  —  1  sin7  3  a:. 

V1    ■   4  1    •   «      \ 

3.  «  =  -( -  sin*  ax  —  -;  snr  aa- )  • 
9       a\i  6  / 

4.  y/  =  i  a;  —  I  sin  4  a?.  5.  y  =  J  a;  —  \  sin  (2  —  4  a;). 

6.  y  =  coss  3  cc (T\  cos3  3  a;  —  \  cos  3  a-). 

7.  y  =  f\  V  cos  2  x  (cos2  2  a;  —  7). 

8.  y  =  i  cos  (2  a  +  1)  [cos2  (2  x  +  1)  -  3]. 

9.  y  =  ^  tan8  a  4-  tan2  x  4-  tan  a;. 

10.  y  =  -  tan3 |  -  2  tan  |  4-  a?. 

11.  y=-£ctn2(a;2  +  a2). 

12.  y  =- -etn5-  +  ctn3-- 3ctn^-a-. 

o  o  o  o 

2       .x 

13.  y  =  -sec6-. 


SJ       j/  2  a;        1 

14.  ^  =  25^860-^1- jjsec2-  +  2jsec4 

15-  ^  =  j(csGbx~  ctn&c).  19.  y  =  sin-1  ^  ~*    ■ 

a;  4-  Z 

16.  y  =  sin-1 2a;.  1  x2  +  3 

•i/o         hn  20«  y  =  7rsin-1- -=•' 

17.  y  =  snr1(2x  —  1).  ^      2  2V3 

,a;-2  t2x  —  3 

18.  y  =  sm-1— 21.  y  =  cos-1 5 


PROBLEMS  213 

3  x  4-  1  ,     |l  —  x 

22.  y^cos-1        ^      •  29.  y  =  ctn-1^-^-- 


23.   y  =  COS-1-^ 


30.  y  =  sec-1  3  a-. 


24.  y  =  tan-1  (a;  —  2) 


a;2  4-  a?  ,  x  + 

31.  y  =  sec-1— — 


25.  y  =  tan"1  Vx2  +  2x.  ^2.  y  =  sec"1  (a;  +  1)  • 

26.  y  =  tan-1-^==-  33     y  =  csc-i(4  ^  +  4  r) 

27.  ?/  =  ctn-1-^-  34.  y  =  csc-1- 


.'• 


28.  y  =  ctn-1 


x2-l 


35.   y==c8c-i-(-a  +  ?J 


36.  y  =  x  sin-1  Vl  —  x2  —  Vl  —  x2. 

37.  y  =  x2  tan-1-  +  a2  ctn-1  -  -f  ox. 

x  a 

1  „         ,- :       3  a2    .      .  x  —  a 

38.  y  =  —  -(x4-3a)V2ax  —  x2  4-  -^-sin-1 

39.  y  =  log(2x2-4x  +  3).  46>   j/_llogVaa  +  ^-« 

40.  y  =  log  Vx2  +  4  x  4-  3.  X 

1         2  x  —  3  •  47.   //  =  log  sin  x. 

41-  •'/  =  12 log 2x4-3 '  48.  y  =  log(sec 3x4-  tan  3 x). 

42.   y J-logSs-A  i^rfn- 

2v3         3x4-V3  49.   //  =  log  — 


43.   y  =  log 


.     x 

1  1  +  sin 


2 
2  tan  x  4- 1 


V3  -  4  x  4-  x2 

44.  y  =  log(3  x  4-  V9x2  +  2).         5°'  -'7  =  log   tan  x  4-  2 

45.  y  =  \  log  (x:!  4-  Vx°  —  «G).  51.  //  =  log  ctn  x  —  esc  2  x. 

52.  y  =  logVar,  +  4  +  ^-j-j. 

53.  y  =  3  Vx7^4  +  log  (a3  +  Vx4-"4f. 


1 ,      1  -  cos  2 


54-  ^=810gl  +  cos2x      4sin22x 

55.  y  =  x  [(log  ox)2  -  2  log  aa  +  2]. 

56.  y  =  log(x2  +  Vx4  —  l)  -  sec-1  x2 


II I 


TRANSCENDENTAL   FUNCTIONS 


57.  y 

58.  y 

59.  y 


2  x  tan-1  2  x  —  log  Vl  +  4  x\ 
|  log  (2  .r2  +  1)  +  V2  tan"1  x  V2. 


60.  ?/  =  x  sec 


a;  tan-1  a« log  Vl  +  <rx~. 

ax log(oaj  +  Vft2x-2  —  l). 


61. 

y  =  e  x. 

62. 

u  —  ^6 

63. 

y  =  etan-^ 

64. 

y^^flF^y 

G5. 

?/  =  -l«co,!I. 

G6. 

y=Ke8*-e-8*)H 

07. 

e» +  ««»+« 

^~c(l  +  loga) 

69.  ?/=— (a2*3—  a-23?)+x* 

J     41ogftv 

X 

70.  ?/  =  2x-a  log  (ea  -f-  1). 


71.  y 


ear(a  sin  mx  —  m  cos mx ) 
a2  +  m2 


73.  y  =  sin-1 


68.  y  =  — g  (aV  —  2ax  +  2).       74.  ?/  =  sin"1  2a; 

ft  6       "I-  & 


75.  7/ 

76.  ?/ 

77.  y 


ex  +  e- 

—   SPWV~ x  


tan- 1 


log  V2  -  2  e*  +  e2*  +  0*  -  1)  ctn" a  (e*  -  1). 

tan-1V^T2^-1°f("  +  1)- 
Vz2  +  2  a; 


78.  vy  =  31og(x2  +  4)  + 

79.  ?/  ==  ft  log 


■"B^l  +  ti 


tan- 1 


ft  +  V(/2 


Vft1 


80.   y^log^ 


Vx  —  Vf 


Vx  +  Vft 


+  tan-1A  - 


81.  y 

82.  j, 

83.  y 

84.  y 

85.  y 

86.  y 


.  a?  +  ft  a      .      i  /  2   i   o — \ 

sec-1 loglsc  +  ft  +  v.r2  +  2  aa). 

a  x  -f  ft 


87.  ?/  =(tanic)aJ! 


:  2  sin-1  V2  ex  -  e2*  +  V 2  e*  -  e2* log (2  -  a?). 

.(Xy.  88-  ^  =  («2  +  0 

=  (tanV?T^. 


PROBLEMS  215 

Find  -~  in  each  of  the  following  cases : 
ax 

.y  n  91.  sin (./•  +  2  y)  +e2x+y  =  0. 

89.  tan-1^  +  xy  =  0.  ^  "" 

92.  ./•"  +  //  =  0. 

90.  sin-1-  +  ^,f-x2  =  0.  no       r 

y  *  93.  //  —  sec  jv/  —  tan  x^  =  0. 

Find  -^  and  -rk  in  each  of  the  following  cases : 
ax         dxr 

94.  ex  +  e'J  =  er  * ".  96.   log ('.'•'-  +  //-)  —  tan-1  -  =  0. 

95.  tan-^+logV7T?=0.  9?'  °OB(*+y)+ooB(*-y)=l. 

y  98.    r'  +  "  =  //'. 

99.  Show  that  the  portion  of  the  tangent  to  the  curve 
a       a  _i_  V«* 


1  a  —  Vr^ 


-V5 


included  between  the  point  of  contacl  and  the  axis  of  y  is  constant. 
(From  this  property  the  curve  is  tailed  the  tractrix.) 

100.  Draw  the  curve  y  =  c~'"  ens//./-,  and  prove  thai  it   is  tangent 
to  the  curve  y  =  e-01  wherever  they  have  a  point  in  common. 

sin.'' 

101.  Draw  the   curve  y  =  }  and   show  that    it   is  tangent    to 

the  curve  y  =  — ,  wherever  they  have  a  point  in  common. 

102.  Find  the  angle  of  intersection  of  the  curves  y  =  sina:  and 

y  =  'eos  .''. 

103.  Find  the  angle  of  intersection  of  the  curves  y  =  sin  ./■  and 


5  +  f  f 


104.  Find  the  angle  of  intersection  of  the  curves  y=  sins  and 
y  =  sin  2  x. 

105.  Find  the  point  of  inflection  of  the  curve  y  —  (x  + 1)  tan-1x. 

106.  Find  the  points  of  inflection  of  the  curve  y  =  e~'~. 

107.  Find  the  point  of  inflection  of  the  curve  y  =  e1-ar. 

108.  Draw  the  curve  y  =  log  tair.r.    Find  a  point  of  inflection  and 
the  slope  at  that  point. 


216  TRANSCENDENTAL  FUNCTIONS 

109.  Prove  that  the  curve 

y  =  \  x  —  §  sin  x  +  j\  sin  2  x 

lias  an  indefinite  number  of  points  of  inflection,  and  that  two  of 
them  lie  between  the  points  for  which  x  =  6  and  x  —  10  respectively. 

Find  the  turning  points  and  the  points  of  inflection  of  the  follow- 
ing curves,  and  draw  the  curves. 

110.  y  =  xe~*\  113-  //  =  sil^- 

, ,..  o  _r  114.  y  =  2  sin  a;  4-  isin2x. 

111.  y  =  xze  x.  J  2 

112.  }/  =  x3e~x.  115'  xi/  =  a2\og-' 

116.  A  tablet  10  ft.  high  is  placed  on  a  wall  so  that  the  bottom 
of  the  tablet  is  8  ft.  from  the  ground.  How  far  from  the  wall  should 
a  person  stand  in  order  that  he  may  see  the  tablet  to  the  best  advan- 
tage ;  that  is,  in  order  that  the  angle  between  the  lines  from  the 
observer's  standpoint  to  the  top  and  the  bottom  of  the  tablet  may 
be  the  greatest  ? 

117.  One  side  of  a  triangle  is  5  ft.,  and  the  opposite  angle  is  40°. 
Find  the  other  angles  of  the  triangle  when  its  area  is  a  maximum. 

118.  Above  the  center  of  a  round  table  is  a  hanging  lamp.  What 
must  be  the  ratio  of  the  height  of  the  lamp  above  the  table  to  the 
radius  of  the  table  in  order  that  the  edge  of  the  table  may  be  most 
brilliantly  lighted,  given  that  the  illumination  varies  inversely  as 
the  square  of  the  distance  and  directly  as  the  cosine  of  the  angle 
of  incidence  ? 

119.  A  weight  P  is  dragged  along  the  ground  by  a  force  F.  If 
the  coefficient  of  friction  is  K,  in  what  direction  should  the  force  be 
applied  to  produce  the  best  result  ? 

120.  An  open  gutter  is  to  be  constructed  of  boards  in  such  a  way 
that  the  bottom  and  the  sides,  measured  on  the  inside,  are  to  be  each 
5  in.  wide,  and  both  sides  are  to  have  the  same  slope.  How  wide 
should  the  gutter  be  across  the  top  in  order  that  its  capacity  may  be 
as  great  as  possible  ? 

121.  A  steel  girder  27  ft.  long  is  to  be  moved  on  rollers  along  a 
passageway  and  into  a  corridor  8  ft.  in  width  at  right  angles  to  the 
passageway.  If  the  horizontal  width  of  the  girder  is  neglected,  how 
wide  must  the  passageway  be  in  order  that  the  girder  may  go  around 
the  corner  ? 


PROBLEMS  217 

122.  Given  that  two  sides  and  the  included  angle  of  a  triangle 
have  at  a  certain  moment  the  values  8  ft.,  12  ft.,  and  30°  respectively, 
and  that  these  quantities  are  changing  at  the  rates  of  4  ft.,  —  3  ft., 
and  12°  per  second  respectively,  what  is  the  area  of  the  triangle  at 
the  given  moment,  and  how  fast  is  it  changing  ? 

123.  A  particle  of  unit  mass  moves  in  a  straight  line  so  that 

irt 
s  =  6  —  5  sin2  —  >  where  t  is  the  time  and  s  the   distance  from  a 

point  0.  Find  when  the  particle  is  moving  forward  and  when 
backward.  Find  also  the  greatest  distance  which  the  particle 
reaches  from  0,  and  the  force  which  acts  upon  it. 

124.  A  motion  of  a  particle  in  a  straight  line  is  expressed  by  the 
equation  s  =  5  —  2cosa£  Express  the  velocity  and  the  acceleration 
at  any  point  in  terms  of  s. 

125.  Two  paxticles  are  moving  in  the  same  straight  line,  and 
their  distances  from  the  fixed  point  0  on  the  line  at  any  time  t  arc 

respectively  x  =  a  cos  ht  and  as'  =  a  COS (  ht  -f-  .,)'   A   and   a    being 

constants.    Find  the  greatest  distance  between  them. 

126.  If  s  =  ae**  +  be~kt,  show  that  the  particle  is  acted  on  by  a 
repulsive  force  which  is  proportional  to  the  distance  from  the  point 
from  which  s  is  measured. 

127.  If  a  particle  moves  so  that 

s  =  e-i«*(a  sin  ht  +  b  cos  ///), 

find  expressions  for  the  velocity  and  the  acceleration.  Hence  show 
that  the  particle  is  acted  on  by  two  forces,  one  proportional  to  the 
distance  from  the  origin  and  the  other  proportional  to  the  velocity. 
Describe  the  motion  of  the  particle. 

128.  A  revolving  light  in  a  lighthouse  \  mi.  offshore  makes  one 
revolution  a  minute.  If  the  line  of  the  shore  is  a  straight  line,  how 
fast  is  the  ray  of  light  moving  along  the  shore  when  it  passes  the 
point  of  the  shore  nearest  to  the  lighthouse  ? 

129.  A,  the  center  of  one  circle,  is  on  a  second  circle  with  center 
at  B.  A  moving  straight  line,  AMN,  intersecting  the  two  circles  at 
M  and  N  respectively,  has  constant  angular  velocity  about  A.  Prove 
that  BN  has  constant  angular  velocity  about  B. 


218  TRANSCENDENTAL  FUNCTIONS 

130.  BC  is  a  rod  a  feet  long,  connected  with  a  piston  rod  at  C,  and 
at  B  with  a  crank  ,  I  /;,  h  feet  long,  revolving  about  A.  Find  C's  velocity 
in  terms  of  AB's  angular  velocity. 

131.  A  body  moves  in  a  plane  so  that  x  =  acos  t-\-b,y  =  asmt  -f-  c, 
where  t  denotes  thne  and  a,  b,  and  c  are  constants.  Find  the  path  of 
the  body,  and  show  that  its  velocity  is  constant. 

132.  The  parametric  equations  of  the  path  of  a  moving  point  are, 
in  terms  of  the  time  t,  x  =  a  cos  Id,  y  =  b  sin  ht,  where  a,  b,  and  'k 
are  constants  and  a  >  b.  Prove  that  the  path  is  an  ellipse.  Find 
the  velocity  of  the  point  in  its  path.  Find  when  the  velocity  is  a 
maximum  and  when  a  minimum. 

133.  A  particle  moves  so  that  x  =  2  cos  t  —  cos  2 1,  y  =  2  sin  t  — 

IT 

sin  2  t,  where  t  is  the  time.   Find  its  velocity  in  its  path  when  t  —  —  • 

134.  If  a  wheel  rolls  with  constant  angular  velocity  on  a  straight 
line,  required  the  velocity  of  any  point  on  its  circumference  ;  also  of 
any  point  on  one  of  the  spokes. 

135.  Prove  that  a  point  on  the  rim  of  the  wheel  of  problem  134 
is  moving  parallel  to  the  straight  line  on  which  the  wheel  rolls,  with 
a  velocity  proportional  to  its  distance  from  OX. 

136.  Show  that  the  highest  point  of  a  wheel  rolling  with  constant 
velocity  on  a  road  moves  twice  as  fast  as  each  of  the  two  points  in  the 
run  whose  distance  from  the  ground  is  half  the  radius  of  the  wheel. 

137.  If  a  wheel  rolls  with  constant  angular  velocity  on  the  cir- 
cumference of  a  fixed  wheel,  find  the  velocity  of  any  point  on  its 
circumference  and  on  its  spoke. 

138.  If  a  string  is  unwound  from  a  circle  with  constant  angular 
velocity,  find  the  velocity  of  the  end  in  the  path  described. 

139.  A  man  walks  along  the  diameter,  200  ft.  in  length,  of  a  semi- 
circular courtyard  at  a  uniform  rate  of  5  ft.  per  second.  How  fast 
will  his  shadow  move  along  the  wall  when  the  rays  of  the  sun  are 
at  right  angles  to  the  diameter  ? 

140.  How  fast  is  the  shadow  in  the  preceding  problem  moving  if 
the  sun's  rays  make  an  angle  a  with  the  diameter  ? 

141.  A  man  walks  across  the  diameter  of  a  circular  courtyard  at 
a  uniform  rate.  A  lamp,  at  one  extremity  of  the  diameter  perpen- 
dicular to  the  one  on  which  he  walks,  throws  his  shadow  on  the  wall. 
Required  the  velocity  of  the  shadow  along  the  wall. 


PROBLEMS  219 

142.  A  ladder  b  feet  long  leans  against  a  side  of  a  house.  Its  foot 
is  drawn  away  in  the  horizontal  direction  at  the  rate  of  a  feet  per 
second.  Find  the  path  described  by  the  center  of  the  ladder  and  the 
velocity  of  the  center  in  its  path. 

143.  Find  ■—■  and  ■—  for  the  curve  x  =  a  (cos  <f>  4-  <f>  sin  <f>), 
y  =  a  (sin  cf>  —  <f>  cos  <f>). 

144.  Find  —  and  -^  for  the  curve  x  =  a  cos3  <£,  y  =  a  sin3  <£. 

145.  Find  -,'    and    , '.,  for  the  curve  x  =  e*  sin  t,  >/  =  e*  cos  t. 

146.  Prove  that  the  logarithmic  spiral  r  =  ea6  cuts  all  radius 
vectors  at   a  constant  angle. 

147.  Prove  that  the  angle  between  the  normal  and  the  radius 
vector  to  any  point  of  the  lemniscate  is  twice  the  angle  made  by 
the  radius  vector  and  the  initial  line. 

148.  Prove  that  the  angle  between  the  cardioid  r  =  a(l  —  cos  0) 
and  a  radius  vector  is  always  half  the  angle  between  the  radius 
vector  and  the  initial  line. 

149.  I£j>  is  the  perpendicular  distance  of  a  tangent  from  the  pole, 
prove  that  p 


^hW 


150.  If  a  straight  line  drawn  through  the  pole  0  perpendicular  to 

a  radius  vector  OP  meets  the  tangenl    in  .1   and  the  normal  in  B, 

show  that  OA  =  r1—-  and  OB  =  —  • 
dr  ad 

151.  Show  that  for  any  curve  in  polar  coordinates  the  maximum 
and  the  minimum  values  of  ;•  occur  in  general  when  the  radius 
vector  is  perpendicular  to  the  tangent. 

152.  Sketch  the  curve  r  =  2  4-  sin  3  6,  and  find  the  angle  at  winch 
it  meets  the  circle  r  =  2. 

153.  Sketch  the  curve  7-2  =  a2sin->  and  determine  the  angle  at 
which  it  intersects  the  initial  line. 

154.  Sketch  the  curves  i~  =  a'2  sin  2  6  and  r2  =  «2cos2  0,  and 
show  that  they   intersect  at  right  angles. 


•2'20  TRANSCENDENTAL  FUNCTIONS 

155.  If  a  particle  traverses  the  cardioid  /•  =  a(l  —  cos  6)  so  that 
6  makes  uniformly  two  revolutions  a  second,  find  the  rate  at  which 
/•  changes,  and  the  velocity  of  the  particle  in  its  path :    (1)  when 

0  =  |  ■   (2)  when  6  =  w. 

156.  Find  the  velocity  of  a  point  moving  in  a  limacpn 

r  =  a  cos  6  +  b 

when  0  changes  uniformly. 

f) 

157.  When  a  point  moves  along  the  curve  r  =  4  sin3  -  at  a  uniform 

o 

rate  of  2  units  per  second,  find  the  rates  at  which  6  and  r  are 
changing :  (1)  when  6  —  — ;  (2)  when  6  =  ir. 

158.  Find  the  radius  of  curvature  of  the  curve  ar  +  y^  =  a*. 

159.  Find  the  radius  of  curvature  of  the  catenary  y  =  -(ea  +  e   a). 

160.  Show  that  the  catenary  y  =  -(ea  +  e  a)  and  the  parabola 

1  ' 

y  =  a  +  —  x1  have  the  same  slope  and  the  same  curvature  at  their 

common  point. 

161.  Find  the  radius  of  curvature  of  the  curve  if  =  £7(x  —  2)3 
at  the  point  for  which  x  —  3. 

162.  Find  the  radius  of  curvature  of  the  cycloid 

i  «  —  x         /T. — ; 5 

y  =  a  cos-1 ±-\2ax  —  x2 

at  a  point  for  which  x  =  —  • 

163.  Find  the  radius  of  curvature  of  the  curve  y  =  e~2xsin  3x 
at  the  origin. 

164.  Find  the  least  radius  of  curvature  of  the  curve  y  =  logx. 

165.  Find  the  points  of  greatest  and  of  least  curvature  of  the 
sine  curve  y  =  sin  x. 

166.  Show  that  the  curvature  of  the  parabola  y  =  ax2  +  hx  +  c 
is  a  maximum  at  the  vertex. 

167.  Show  that  the  product  of  the  radii  of  curvature  of  the  curve 

y  =  ae  a  at  the  two  points  for  which  x  =  ±  a  is  a2(e  +  e~l)z. 


PEOBLEMS  221 

168.  Find  the  radius  of  curvature  of  the  four-cusped  hypocycloid 
x  =  a  cos3</>,  y  =  a  sin3<£. 

169.  By  use  of  the  parametric  equations  of  the  ellipse  find  the 
points  where  the  radius  of  curvature  is  a  maximum  or  a  minimum, 
and  the  values  of  these  radii. 

170.  Find  the  radius  of  curvature  of  r  =  a  (2  cos  0  —  1). 

171.  Find  the  radius  of  curvature  of  the  lemniscate  ia  =  2  aa  cos  2  0. 

172.  Find  the  greatest  and  the  least  values  of  the  radius  of 
curvature  of  the  curve  r  =  a  sin3  -  • 

173.  If  the  angle  between  the  straight  line  drawn  from  the  origin 
perpendicular  to  any  tangent  to  a  curve  and  the  radius  vector  to 
the  point  of  contact  of  the  tangent  is  either  a  maximum  or  a  mini- 

r* 

mum,  prove  that  p  =  — ,  where  p  is  the  length  of  the  perpendicular. 


CHAPTER  XII 
INTEGRATION 

109.  Introduction.  In  §  80  the  process  of  integration  was 
defined  as  the  determination  of  a  function  when  its  derivative  or 

its  differential  is  known,  and  was  denoted  by  the  symbol  I  ; 

that  is,  if  „,  N  7         ,„,  N  J 

f(x)dx  =  dF(x), 

then  ff(x)  dx  =  F(x).  (1) 

The  expression  f(x)  dx  is  said  to  be  under  the  sign  of  inte- 
gration, f(x)  is  called  the  integrand,  and  F(x)  is  called  the 
integral  of  f(x)dx;  sometimes  F(x)  is  called  the  indefinite 
integral,  to  distinguish  it  from  the  definite  integral  defined 
in    §  81. 

The  determination  of  the  indefinite  integral  is  important  in 
a  wide  range  of  problems,  and  for  that  reason  we  shall  now 
deduce  formulas  of  integration. 

We  ought  to  note  first,  however,  that  a  more  general  form 
of  (1)  is  p 

jf(x)dx  =  F(x)+C,  (2) 

where  C  is  the  constant  of  integration  (§80).  In  each  of  the 
formulas  we  shall  derive,  C  will  be  omitted,  since  it  is  inde- 
pendent of  the  form  of  the  integrand,  but  it  must  be  added 
in  all  the  indefinite  integrals  determined  by  means  of  them. 

110.  Fundamental  formulas.    The  two  formulas 

fcdu  =  cfdu  (1) 

and      j(du  +  dv  +  dw  -\ )  =  fdu  +  fdv  +  fdw  -J (2) 

are  of  fundamental  importance,  one  or  both  of  them  being  used 

222 


INTEGRAL  OF  U"  223 

in  the   course    of   almost  every  integration.     Stated   in   words 
they  are  as  follows: 

(1)  A  constant  factor  may  be  changed  from  one  side  of  the  sign 
of  integration  to  the  other. 

(2)  The  integral  of  the  sum  of  a  finite  number  of  functions  is 
the  sum  of  the  integrals  of  the  separate  functions. 

To  prove  (1),  we  note  that  since  cdu  =  d(cu),  it  follows  that 


I  cdu  =  I  d(cu)  =  cu  =  <?  I  < 


///. 


In  like  manner,  to  prove  (2),  since 

du  +  dv  +  dw  H =  d(u  +v+w+ •  ■  •)» 

we  have 

j  (du  +  dv  +  dw  H )  =  jd(u  +  v  +  w  -\ ) 

=  u  +  v  +  w  +  •  ■  • 

=  Cdu+  Cdv  +  Cdw-\ . 

The  application  of  these  formulas  is  illustrated  in  the  follow- 
ing articles. 

111.  Integral  of  un.    Since  for  all  values  of  m  except  m—  0, 
d{  //m)  =  mum~] thi, 


it  follows  that 

Placing  m  =  n  +  l,  we  have 


d\      )=  it'"   ldi(, 
\m/ 

C     m-l    7              "'" 
I    U  du  = 

J  m 


J 


<-„T>  <" 

for  all  values  of  n  except  n  =  — 1. 

In  the  case  n  =  —  1,  the  expression  under  the  sign  of  inte- 
gration in  (1)  becomes  — ,  which  is  recognized  as  c?(log?<). 

Therefore  /  - —  =  logw.  (2) 


224  INTEGRATION 

In  applying  these  formulas  the  problem  is  to  choose  for  u 
some  function  of  x  which  will  bring  the  given  integral,  if  pos- 
sible, under  one  of  the  formulas.  The  form  of  the  integrand 
often  suggests  the  function  of  x  which  should  be  chosen  for  u. 

Ex.  1.    Find  the  value  of  f(ax*+  bx  +-+  ^)dx. 

Applying  (2),  §  110,  and  then  (1),  §  110,  we  have 

flax*  +  bx  +  -+  ^Sdx  =  fax2dx+fbxdx  +f-/lx  +  f~2dx 

=  afx*dx  +  bjxdx  +  cf—  +  efx-*dx. 

The  first,  the  second,  and  the  fourth  of  these  integrals  may  be  evaluated 
by  formula  (1)  and  the  third  by  formula  (2),  where  u  =  x,  the  results  being 

11  e 

respectively  -  ax3,  -  bx2, ,  and  c  log  x. 

Therefore 

f(ax2  +  bx  +  -  +  ~)dx  =  -  axs  +  -bx*  +  c\ogx--+  C. 
J  \  x      x-j  3  2  x 

•s/ 
Ex.  2.    Find  the  value  of  f(.r2+  2)xdx. 

If  the  factors  of  the  integrand-  are  multiplied  together,  we  have 

f(x2  +  2)  xdx  =f(x8  +  2  x)  dx, 

which  may  be  evaluated  by  the  same  method  as  that  used  in  Ex.  1,  the 
result  being  \  x*  +  x2  +  C. 

Or  we  may  let  x2  +  2  =  u,  whence  2  xdx  =  du,  so  that  xdx  =  \  du.   Hence 


f(x2+  2)xdx  =j\udu  =  ljudu 


2    2 
=  i  (a*  +  2)2  +  C. 

Instead  of  actually  writing  out  the  integral  in  terms  of  u,  we  may  note 
that  xdx  =  ^d (x2  +  2)  and  proceed  as  follows : 

f  (x2  +  2)  x  dx  =  f  (x2  +  2)  1  d  (x2  +  2) 

.=  }jf(x2+2)d(x2+2) 

=  \(x2+2y+C. 
Comparing  the  two  values  of  the  integral  found  by  the  two  methods  of 
integration,  we  see  that  they  differ  only  by  the  constant  unity,  which  may 
be  made  a  part  of  the  constant  of  integration. 


INTEGRAL  OF  UN  225 

Ex.  3.    Find  the  value  of    C  (ax2  +  2  bx)s(ax  +  b)dx. 

Let  ax-  +  2  bx  =  u.   Then  (2  ax  +  2  b)  dx  =  du,  so  that  (ax  +  b)  dx  =  I  du. 

Hence       f  (ax2  +  2  bxf  (ax  +  b)  dx  =   f  \  u3  du  ^ 

=  \  (ax1  +  2  bxy  +  C. 

Or  the  last  part  of  the  work  may  be  arranged  as  follows : 

f  (ax*  +  2  bx)z  (ax  +  b)  dx  =   f  (ax2  +  2  bx)*  \  d  (ax*  +  2  6a:) 

=  I  f(ax2  +  2  bxf  d  (ax2  +  2  6x) 

=  I  (ax2  +  2  bx)*  +  C. 

Ex.4.    Find  the  value  of    f  *  C*  +  V'** . 
J      ax2  +  2  bx 

As   in    Ex*.  3,   let   ax2  +  2  bx  =  u.     Then    (2  ax  +  2  b)dx  =  du,   so   that 

(ax  +  b)  dx  =  I  du. 


Hence 


/4  (ax  +  b)  dx  _    r  2  <lu  _  0  r  du 
ax2  +  2  bx    ~  J      u     ~  ~  J     u 


=  2  log  u  +  C 

=  2  log  (ar-  +  2  bx)  +  C 

=  log  (ax8  +  2&r)3+  <\- 


J 

ax2  +  2  bx         J        ax2  +  2  bx 

_  .,   f  ' '  ("S2  +  2  bx) 
~    J       ax2  +  2bx 

=  2  log  (OX8  +  26x)  +  C 

=  log(oxa  +  2//./)2+  C. 

Ex.  5. 

Find  the  value  of    f  (eaj'  +  />)-<"'  dx. 

Let  t"J 

+  b=  u. 

Then 

>'■'  ad.r  = 

du. 

Hence 

/<- 

•+/,)-,"■ 

■dx=  r+i* 
j  ■  <i 

a  ^ 

C(eax  +  b)2tn' 

-,/.i-  =    f  _  (,."<■  +  b)2d(ea'-  +  6) 
J    a 

=  1   f(^  +  b)2d  (e°*  +  6) 

J_  (gax  +  6)8  +  C. 


226  INTEGRATION 

_     ,    p.    ,  ,,         ,        ,    P  sec2 (ax  +  b)dx. 
Ex.  6.    Find  the  value  of   |  i - 

J   tan  (ax  +  b)  +  c 

Let  tan  (ax  +  /;)  +  c  =  w.    Then  sec2  (ox  +  b)adx  =  du. 

Hence 

•v   tan  (owe  +  b)  +  c      *>    a 

du 


/sec2 (ax  +  />)ox  _    r  1   c/m 
tan  (ax  +  b)  +  c      J    a     u 

1    rdu 
a  J     u 


-logu  +  C 


-  log  [tan  (ax  +  b)  +  c]  +  C, 


/•  sec2  (ax  +  V)dx  _  rl    d  [tan  (ax  +  b)  +  c] 
«/   tan  (ax  +  6)  +  c      J   a       tan  (ax  +  b)  +  c 

_  1   /•  </[tan  (ax  +  fr)  +  c] 
a  «/       tan  (ax  +  6)  +  c 

=  -  log  [tan  (ax  +  b)  +  c]  +  C. 

The  student  is  advised  to  use  more  and  more  the  second 
method  illustrated  in  the  preceding  problems  as  he  acquires 
facility  in   integration. 

112.  Integrals  of  trigonometric  functions.  By  rewriting  the 
formulas  (§  96)  for  the  differentiation  of  the  trigonometric  func- 
tions we  derive  the  formulas 

I  cos  udu  =  sin  u,  (1) 

I  sin  udu  =  —  cos  u,  (2) 


/ 


seG2icdu  =  tan  u,  (3) 

esc'2  u  du  =  —  ctn  u7  (4) 


I  sec  u  tan  udu  =  sec  u,  (5) 

/  esc  u  ctn  w  Jw  =  —  esc  u.  (6) 


/ 


TRIGONOMETRIC  FUNCTIONS  227 

In  addition  to  the  above  are  the  four  following  formulas: 

tan  udu  =  log  sec  u,  (7) 

ctn  udu  — log  sin.  u,  (8) 

|  secwc?w  =  log(secw  +  tanw)=  log  tan(-  +  -J»         (9) 

it 
esc  udu  =  log  (esc  u  —  ctn  u)  =  log  tan  -  •  (10) 

To  derive  (7)  we  note  that  tan  u  =  — : — ,  and  that  —  sinwcJw 
=  d(cosu).    Then 

I  tan  u  dit  —  —  |   — - 

J  J       cos  u 

=  —  log  COS  u 

=  log  sec  u. 

In  like  manner,         /  ctmidu  =  I  — =  log  sin  w. 

J  J      sin  u 

Direct  proofs  of  (9)  and  (10)  will  not  be  given  here.  At 
present  they  may  be  verified  by  differentiation.  For  example, 
(9)  is  evidently  true  since 

d  log  (set-  u  +  tan  u)  =  see  u  <hi. 

The  second  form  of  the  integral  may  be  found  by  making  a 
trigonometric  trans  formation  of  sec  u  +*tan  u  to  tanl  — +  -)• 

Formula  (10)  may  be  treated  in  the  same  manner. 

Ex.  1.    Find  the  value  of  |cos(ara  +  bx)(2ax  +  b)dx. 

Let  ax*  +  bx  =  u.   Then  (2  ax  +  b)  dx  =  du. 

Therefore     f  cos  (ax2  +  bx)  (2  ax  +  b)  dx  =  J*cos  (ax-  +  bx)  d  (ax2  +  bx) 

=  Bin  (ax2  +  >'■>)  +  C. 
Ex.  2.    Find  the  value  of  f  sec  (ea3?  +  b)  tan  (e**  +  b)  e^xdx. 
Let  e03?  +  b  =  u.    Then  eax22  axdx  =  du. 
Therefore  J^sec  (e**2  +  6)  tan  (e^  +  6)  ea*\r  dx 

=  -L  J  sec  (c«*  +  &)  tan  (ca'3  +  b)  d ( e**'  +  h) 

=  J_  sec  ((•«•' 2+  &)+  C. 


228  INTEGRATION 

It  is  often  possible  to  integrate  a  trigonometric  expression  by 
means  of  formulas  (1)  and  (2)  of  §  111.  This  may  happen  when 
the  integrand  can  be  expressed  in  terms  of  one  of  the  elementary 
trigonometric  functions,  the  expression  being  multiplied  by  the 
differential  of  that  function.  For  instance,  the  expression  to  be 
integrated  may  consist  of  a  function  of  sin  x  multiplied  by  cos  xdx, 
or  of  a  function  of  cosrr  multiplied  by  (—  sin  xdx),  etc. 

Ex.  3.    Find  the  value  of  /  Vsina:  cos3 xdx. 

Since  d  (sin  x)  =  cos  xdx,  we  will  separate  out  the  factor  cos  xdx  and 
express  the  rest  of  the  integrand  in  terms  of  sin  x. 

Thus  Vsina;  coszxdx  =  vsina;(l—  sin2x)  (cos  xdx). 

Now  place  sin  x  =  u,  and  we  have 

I  Vsin  x  cossxdx  =  J  u?  (1  —  u")  du 
=  I  («2  _  u^)du 
=  $  «t  -  f  u%  +  C 
=  ^T  sinlx  (7-  3  sin2x)  +  C. 

Ex.  4.    Find  the  value  of  f  sec6 2  xdx. 

Since  d(tan  2  x)  =  2  sec2 2  xdx,  we  separate  out  the  factor  sec2 2  xdx  and 
try  to  express  the  rest  of  the  integrand  in  terms  of  tan  2  x. 

Thus  sec6 2  xdx  =  sec4 2  a:  (sec2 2  xdx) 

=  (1  +  tan2  2  x)2(sec22xdx) 

=  (1  +  2  tan2 2  x  +  tan4 2  x)  (sec2 2  xdx). 

Now  place  tan  2  x  =  u,  and  we  have 

f sec62  xdx  =  £  f  (1  +  2  m2  +  u4)  du 

=  Jr  tan  2  x  +  ^tan32  x  +  TVtan52  x  +  C. 
Ex.  5.  Find  the  value  of    I  tan5  xdx. 

Placing  tan5x  =  tan3 a;  tan2x  =  tan3 a;  (sec2x  —  1), 

we  have  /  tan5 xdx  =  f  tan3 x sec2 x dx  —  f  tansxdx 

=  ^tan4x  —  |  tansxdx. 


TRIGONOMETRIC  FUNCTIONS  229 

Again,  placing  tan3x  =  tanx  (sec2x  —  1), 

we  have  J  tan3xdx  —  f  tanx  sec2  xdx  —  J  tanxtfx 

=  i  tan2x  +  log  cos  x  +  C. 
Hence,  by  substitution, 

/  tan5x  dx  =  ^  tan4 a:  —  i  tan2  a;  —  log  cos  x  +  C. 

When  the  above  method  fails,  the  integral  can  often  be 
brought  under  one  or  more  of  the  fundamental  formulas  by  a 
trigonometric  transformation. 

Ex.  6.   Find  the  value  of  j  con- xdx. 
Since  cos2x  =  J  (1  +  cos  2  x), 

we  have  i  cos'2  xdx  =  \   f  (1  +  cos  2  x)  dx 

=  ■•  f  lLr+  lfcoa2xd(2z) 

=  l.r+  \  sin  •_'.;  +  C. 

Ex.  7.    Find  the  value  of   /  Bin'z  C0B*xdx. 
Placing  sin2x  cos4x  =  (sin  x  cos  ar)acos8  v, 

we  have  sin^x  cos4x  =  |  sin2  2  x(l  +  cos  2  x). 

Therefore        J  sin2x  cos4xr/x  =  \  j  sin-2xdx  +  l  I  sin2 2  x  cos2xtfx. 
Using  the  method  of  Ex.  6,  we  have 

|  sin-  2  x  dx  —  \  f  (1  —  cos  4  x)  ilx 

=  h x  ~  \ sin  1  •''• 
Writing         sin2  2  x  cos  2  x  dx  =  sin2  2  x  (cos  2  x  dx) 
and  placing  sin  2  x  =  u,  we  have 

/  sin2 2  x  cos  2  xdx  =  £  j  u9<£« 

=  J  sin3  2  x. 
Combining  these  results,  we  have,  finally, 

f  sin2x  cos4x</x  =  rV  x  +  ?V  s*n3  -  x  —  sV  s*n4  x  +  C. 


230  INTEGRATION 

Ex.  8.    Find  the  value  of    J  Vl  +  cosxdx. 
Since  cos  x  =  2  cos-  - 


Therefore 


113.  Integrals  leading  to  inverse  trigonometric  functions.  From 
the  formulas  (§97)  for  the  differentiation  of  the  inverse  trig- 
onometric functions  we  derive  the  following  corresponding 
formulas  of  integration : 


/ 


VI  -  u 
lu 


i 


/. 


1  +  w 
du 


=  sin  'm  or  -  cos-1w, 
=  tan-1?/ or  —  ctn-1w, 


sec     u  or 


mVm2  —  1 

These  formulas  are  much  more  serviceable,  however,  if  u  is 

u 
replaced  by  -  (a>  0).     Making  this  substitution   and  evident 

reductions,  we  have  as  our  required  formulas 

du  ,  u  _, N 

=  sin  1  - ,  (1) 


/ 


Va2  -  u2  a 


f 


du  1        _1u  .0. 

tan  l-,  (2) 


a  -f-  w       a  a 


/; 


^  1  ,  ?*  ,ON 

sec     -.  (o) 


Vm2  -  a.2      «  « 

Only  one   of  the   possible   values  has  been  given  for  each 
integral,  as  that  single  value  is  sufficient  for  all  work. 

Referring  to  1,  §  97,  we  see  that  sin-1-  must  be  taken  in  the 

first   or  the   fourth   quadrant ;    if,  however,  it  is  necessary  to 


INVERSE  TRIGONOMETRIC  FUNCTIONS  231 

have  sin-1-  in   the  second  or  the  third   quadrant,  the   minus 

a 

sign  must  be  prefixed.    In  like  manner,  in  (3),  sec-1-  must  be 

taken  in  the  first  or  the  third  quadrant  or  else  its  sign  must 
be  changed. 

J   y/9  -  4  , 


Ex.  1.    Find  the  value 

x2 


-  -sin-1 h  C. 


Letting  2x  =  u,  we  have  du  =  2  dx,  and 

f      dx       =1  r    ^(^Q 
J  V !)  -  4  x-      2  J  A  9  _  (O  x)2      - 

/rfa; 
—  • 

z  V  3  a:2  —  4 

If  we  let  V3a:  =  u,  then  ^/u  =  v3  da?,  and  we  may  write 

f       dx  r         fl(VHx)      _  =  1scc-i^a? 

^a;V3^-4     J  V3*- V(V3*)2- *     2  2 

—  • 

\  1  ,■  -  x* 

Since  \'l.c-r  =  v/l  -  (.<•  -  'J)-, 

,  have    r-j^=  =  r     rf*      =  r  ^*-2> 

J  VI  x  -  x-     ^1-  (,■  -  2)-     J  \  1  -    (./■  -  2)2 

x  —  2 

=  sin"1"    t)  "  +  C. 

/ill' 
— 
2  ./•-  +  3  x  +  5 

To  avoid  fractious  and  radicals,  we  place 

dx  8  -A--  n  1  </<■ 


+  C. 


2  a •-  +  :!  x  +  5      16  x-  +  24  r  +  40  (4  x  +  ■'>)-  +  31 

Therefore 

f         dx  =     r         4  dx  =  2  r    rf(^  +  3) 

^2x2+3a-  +  5        J  (4a:  +  3)a+31      "J  (4  a:  +  3)2  +  31 

=  — — =tan_1  — +  C. 

V31  V31 

The  methods  used  in  Exs.  3  and  4  are  often  of  value  in  dealing  with 
functions  involving  ax2  +  hx  +  c. 


232  INTEGRATION 

Ex.  5.    Find  the  value  of    f  (?+*)&. 
Separating  the  integrand  into  two  fractions,  that  is, 


5  +  1  x4      5  +  4  x4 
and  using  (2),  §  110,  we  have 

/(x3  +  x) dx  _  r    x8 dx         r    xdx 
5  +  4x4    ~J  5  +  4x4     J  5  +  4x4' 

■d  4.  r  xZ(Ix      i  r  i6x3f/x     i  .    ,_  ,  .  .x 

But  i^M^  =  IfiJ.^M^  =  Wl0g(5  +  4a:)' 

and  f    *^     =1  f     **«*      =J-tan-i^. 

J5  +  4x4      4  J  5 +  (2*3)3      4V5  V5 


Therefore 

/■  (x3  +  x)  r/x       1  .       „   ,    .    4N  ,       1     ,         ,  2  x2  ,    _ 

I  H 7^7-  =  77:  1o8' (5  +  4 x4)  + —tan-1  — -  +  C. 

J      5  +  4x4        10     bV  wJ      4Vo  V5 

Ex.  6.    f^Y^~2  =  P*11""1*]^?  =  tan_1  ^  -  tan-i  (-  1). 

There  is  here  a  certain  ambiguity,  since  tan_1V3  and  tan-1(  —  1)  have 
each  an  infinite  number  of  values.  If,  however,  we  remember  that  the  graph 
of  tan_1x  is  composed  of  an  infinite  number  of  distinct  parts,  or  branches,  the 
ambiguity  is  removed  by  taking  the  values  of  tan-1  v3  and  tan_1(—  1)  from 

=  tan-1  b  — 

..  1  +  x2 
tan-1  a  and  select  any  value  of  tan-1o,  then  if  b  =  a,  tan-1  J  must  be  taken 
equal  to  tan-1  a,  since  the  value  of  the  integral  is  then  zero.    As  b  varies 
from  equality  with  a  to  its  final  value,  tan_1i  will  vary  from  tan-1  a  to  the 
nearest  value  of  tan-1^. 

The  simplest  way  to  choose  the  proper  values  of  tan_1&  and  tan-1  a  is 

to  take  them  both  between and  —  •    Then  we  have 


/ 


V3    dx     _ir      I     7r\  _  7  tt 
-l  1  +  x2  ~  3      \      4/      12 ' 


Ex.  7. 


J2  —  =    sin-1-      =  sin-1  \  —  sin-10. 

0   Va2-x2      L  «Jo 


The  ambiguity  in  the  values  of  sin-1  \  and  sin-10  is  removed  by  notic- 
ing that  sin-1-  must  lie  in  the  fourth  or  the  first  quadrant  and  that  the 

a 
two  values  must  be  so  chosen  that  one  comes  out  of  the  other  by  continuous 
change.    The  simplest  way  to  accomplish  this  is  to  take  both  sin-1  \  and 

sin-10  between and  —  • 

2  2 


Then 


/»2       dx 

Jo   Va2- 


LOGARITHMIC  FUNCTIONS  233 

114.  Closely  resembling  formulas   (1)   and   (2)  of  the   last 
article  in  the  form  of  the  integrand  are  the  following  formulas : 

f-r^-;  =  l0g(m-V^T7»)>  (1) 

J   V  u'-  +  a- 

f-^L=  =  \og(u+V^^),  (2) 

J  Vtt3  —  aA 

,  C    du  1   ,      a  —  a         1   .      a  —  u       .0. 

and  I  — -  =  -— log or-— log —  •     (3) 

Jir-a-      2  a     au  +  a       2  a     sa  +  u      w 

To  derive  (1)  we  place  u  =  a  tan  <f>.   Then  du  =  a  sec2<£c/$,  and 
W2+  a'2  =  a  sec  </>.    Therefore 

/,  =  I  sec  &J& 

Vw2+a2    J 

=  leg  (sec  <ft  +  tancft)  (by  (9),  §  112) 
?/.+vV+aa 


=lo4 — 5 


=  log  (if  -\-Vu*+  a')—  log  a. 

But  log  a  is  a  constant  and  may  accordingly  be  omitted  from 
the  formula  of  integration.  If  retained,  it  would  affect  the 
constant  of  integration  only. 

To  derive  (2)  we  place  u  =  a  sec  <f>  and  proceed  as  in  the 
derivation  of   (1). 

Formula  (3)  is  derived  by  means  of  the  fact  that  the  fraction 

— may  be  separated  into  two  fractions,  the  denominators 

of  which  are  respectively  u  —  a  and  u  +  a ;  that  is, 
1  1/1  1 

a\u  —  a      u  +  a 


I      ri     1  1     \ 

du 

hi  r  du 


Then  fL-**  r/_L__u 

J  W  —  a'      2  a  J  \u  —  a      u  +  a) 
_!_{  P   du  C   du 

2  a  \  J  u  -  a      J  u  +  a 


=  2^l0g  (M  _  ^  ~  l0g (M  +  ^ ^ 

1   .      u  —  a 

=  ——  log 

2a     bu+a 


234  INTEGRATION 

The  second  form  of  (2)  is  derived  by  noting  that 

r  du        r  -  du    ,     . 

I  =    /  =  log(«  —  «•). 

The  two  results  differ  only  by  a  constant,  for 


a  +  u  u-\-  a 

and  hence  log =  log(—  1)  +  log , 

a  +  u  u  +  a 

and  log(— 1)  is  a  constant  complex  quantity  which  can  be  ex- 
pressed in  terms  of  V—  1. 

dx 


Ex.  1.    Find  the  value  of    f  - 


V 3  x2  +  4  x 

To    avoid    fractions   we   multiply  both   numerator    and    denominator 
by   V3. 

dx  _        V3  dx  -y/Srlx 

Then  V3  x2  +  4  x       V9  a;2  +  12  x       V(3  a;  +  2)2  -  4 


Letting  3  x  +  2  =  w,  we  have  c/w  =  3  tfx,  and 

r         dx  _    1  -Jr 3  r/x 

■'  V«  a;2  +  4  x       -n/3^  V(3  x  +  2)2  -  4 


=  ^-log  (3  x  +  2  +  V(3  x  +  2)2  -  4 )  +  C 
V3 

=  -^=log  (3  x  +  2  +  V9x2  +  12x)  +  C. 
V3 

/<7x 

2  x2  +  x  -  15 

Multiplying  the  numerator  and  the  denominator  by  8,  we  have 

-       r  dx  _  2  r  4  dx 

J  2  x2  +  x  -  15       "J  (4  x  +  1)'~  -  (ll)2 

=  1_        (4x  +  l)-ll 

11     &(4x  +  l)  +  ll  +  L- 

1         2  x  —  5  1         2  a;  —  5       1 

This  may  be  reduced  to  —-log +  C,  or  —-log - —  — -  log  2  +  C, 

11        2x  +  b  11         x  +  3        11 

and  the  term  —  ^  log  2,  being  independent  of  x,  may  be  omitted,  as  it  will 
only  affect  the  value  of  the  constant  of  integration. 

Ex.  3.    Find  the  value  of    (  -± '- 

J  2  x2  +  x  -  15 

If  2  x2  4-  x  -  15  =  u,  du  =  (4  x  +  1)  dx. 

Now  3  x  +  4  may  be  written  as  J  (4  x  +  1)  +  ^-. 


LOGARITHMIC  FUNCTIONS  235 

Therefore       f  (3*  +  4)<fa  =  f  »(** +  1)  +  W* 
J  2  x2  +  x  -  15      J  2  x2  +  x  -  15 


_  3    r  (4x  +  l)r/x        13  /»  dx 

~  4  J  2  x-  +  x  -  15       4  J  2  x2  4-  x 


15 


The  first  integral  is  §  log  (2  x2  +  x  - 15),  by  (2),  §  111,  and  the  last 

13        o  ,-  _  ^ 
integral  is  of  the  form  solved  in  Ex.  2  and  is  —  log 


Hence  the  complete  integral  is 

3log(2x2  +  x-15)  +  Hlog2x 


4     °  v  '      44     "  x  +  3 

(2x4-5)  dx 


Ex.  4.    Find  the  value  of    f 


V 3  x2  +  4  x 

The  value  of  this  integral  may  be  made  to  depend  upon  that  of  Ex.  1 
in  the  same  way  that  the  solution  of  Ex.  3  was  made  to  depend  upon  the 
solution  of  Ex.  2.    For  let  3  x2  +  4i  =  u;  then  du  =  (0  z  4  1 ) dx. 

Now  2  x  +  5  =  J  (-6  x  +  1)  +  ' .' . 


Therefore 


r(2x  +  5)dx  _  r  [\(6x  +  4)  +  Jnv]^x 
J    V3x2+4x     ^  V3  x2  +  4  x 

=  \   f  (3  x2  +  4  x)- 1  [(6  x  +  4)  rfx]  +  ^  f         **  • 

^  J  ^J  V3  x2  +  4  x 

The  first  integral  is  ?,  V3  x2  +  4  x,  by  (1),  §  111,  and  the  Becond  inte- 
gral is  =  log  (3  x  +  2  +  ">/$  x2  +  12 x),  by    Ex.1.    Hence  the  complete 

3  V3 

integral  is 


11 


£  V3  x2  +  4  x  4-  — ^plog  (3x42  +  V'.»x2  +  12x)  4-  C. 
y  3  V  3 


Ex.5.    Find  the  value  of    jsecxdx. 

/r  dx         r  coQxdx 
secxdx  =  I  =  I 
J  cosx      J 


cosx      t/      cos'x 

J  sm-  x  -  1  2     &  1  +  sin  x 

1  .      1  +  sin  x 
-log- : —  4-  C 

2  1  —  sin  x 


1,      (1  +  sinx)2       ~ 
2IOg    l-sin2x    +C 

ilog(l± 

2     °\    c< 


sinx\2      _ 


2        \    cos  x 
=  log  (sec  x  +  tan  x)  4-  C. 


236  INTEGRATION 

<t.r 


Ex.  6.    Find  the  value  of 


f— ~ 

J  1  +  2 


+   1  COS  X 


As  in  Ex.  8,  §  112,  we  place  cosx  =  2 cos2-  —  1. 


Then  .1  +  2  cos  x  —  4  cos2  -  —  1, 


and 


dx  r        dx 


/dx         _  r 
1  +  2  cos  x  ~  J 


4  cos2-  —  1 


Multiplying  both  numerator  and  denominator  by  sec2  - ,  we  have 


Now  let  tan  -  =  z.     Then   sec2 -dx  =  2  dz,  and   the  integral   assumes 
the  form 


2dz  2     .      g-V3 

— -  log  — 

V3        z  +V3 

tan  -  +  V3 


9 


Hence  f        ^ =  — -  log +  C. 

Jl  +  2cosx       V3     %an|-V3 


115.  Integrals  of  exponential  functions.    The  formulas 

eudu  =  eu  (1) 

and 


Caudu  =  -^-au  (2) 

J  log  a 

are  derived   immediately  from  the  corresponding  formulas  of 
differentiation.    The  proof  is  left  to  the  student. 
116.  Collected  formulas. 


I 


lu  =  ^—-,  (1) 

^  =  logW,  (2) 


FORMULAS  237 

I  cos  u  du  =  sin  u,  (3) 

/  sin  u  du  =  —  cos  u,  (4) 

J  sec2  u  du  =  tan  u,  (5) 

I  csc2w  c?m  =  —  ctn  w,  (6) 

I  sec  it  tan  w  du  =  sec  m,  (7) 

I  esc  w  ctn  udu  =  —  esc  w,  (8) 

|  tan  udu  =  log  sec  w,  (9) 

I  ctn  udu  =  log  sin  a,  (10) 
/  sec  it  du  =  log  (sec  ?<  +  tan  ?<)  =  log  tan  (  -  -f  -  j ,      (11) 

/  esc  u  du  =  log  (est'  n  —  ctn  u~)  =  log  tan  -  ,  (  1  J  ) 

/d!ie  ,  u 

-.             =sin-1-t  (13) 

- ^  =  -tan"1-,  (14) 

a2  +  u2      a  a 

\  —             =  -  sec     -,  (15) 
J  MvV-aa      a           « 

f-y==  -  log(«  +  V5F=rf>  (17) 

J  Vu~  —  a 

/du           1   ,      u  —  a         1   .      a  —  u  .+  ^ 

-j =  =  —  log  — ■ —  or  —  log ,  (18) 
m2  —  a2      2a     °  w  +  a        2a     °  a  +  u 

fe"du  =  eu,  (19) 

CaHdu  =  - a".  (2(T) 

J                 log  a  v 


238  INTEGRATION 

117.  Integration  by  substitution.  In  order  to  evaluate  a  given 
integral  it  is  necessary  to  reduce  it  to  one  of  the  foregoing 
standard  forms.  A  very  important  method  by  which  this  may 
be  done  is  that  of  the  substitution  of  a  new  variable.  In  fact, 
the  work  thus  far  has  been  of  this  nature,  in  that  by  inspection 
we  have  taken  some  function  of  x  as  u, 

In  many  cases  where  the  substitution  is  not  so  obvious  as  in 
the  previous  examples,  it  is  still  possible  by  the  proper  choice 
of  a  new  variable  to  reduce  the  integral  to  a  known  form. 
The  choice  of  the  new  variable  depends  largely  upon  the  skill 
and  the  experience  of  the  worker,  and  no  rules  can  be  given 
to  cover  all  cases.  We  shall,  however,  suggest  a  few  substi- 
tutions which  it  is  desirable  to  try  in  the  cases  denned. 

I.  Integrand  involving  fractional  'powers  of  a  +  bx.  The  substi- 
tution of  a  power  of  z  for  a  +  bx  will  rationalize  the  expression. 

x-ilx 
(1  +  2  x)i 
Here  we  let  1  +  2  x  =  z3 ;  then  x  —  h  (~3  —.1)  and  dx  =  %  z2dz. 


Ex.  1.    Find  the  value  of   I 


Therefore  f — — —  =  -  f  (z'  -  2  z4  +  z)dz 

J  (1+  2x)i      SJ 

=  3io-2(5~6-16z3  +  20)  +  <?. 
Replacing  z  by  its  value  (1  +  2  x)  ~s  and  simplifying,  we  have 


/ 


x'2dx  3 


(1+2*)*      320 


(1+2  x)i  (9  -  12  x  +  20  a,-2)  +  C. 


II.  Integrand  involving  fractional  powers  of  a  -\-bxn.  The  sub- 
stitution of  some  power  of  z  for  a  +  bx11  may  rationalize  the 
expression.  

Ex.  2.    Find  the  value  of  I  — ■ dx. 

J       .  x 

We  may  write  the  integral  in  the  form 


./' 


i (xdx) 


and  place  x2  +  a2  =  z2.    Then  xdx  =  zdz,  and  the  integral  becomes 
J  z2  -  ,r      J   \        z2-a2/  2     &  z  +  a 


SUBSTITUTION  239 

Replacing  z  by  its  value  in  terms  of  x,  we  have 


/Vx2  +  a~  .          /-=— — -;  ,   a  ,       Vx2  +  a-  —  a  ,   „ 
etc  =  V a;2  +  «2  +  -  log     /  +  C. 

x  2         Vx2  +  a*  +  a 


Ex.  3.    Find  the  value  of  JV(1  +  2  x3)^/x. 
We  may  write  the  integral  in  the  form 

|V(1  +  2x3)2(xVx), 
and  place  1  +  2  x3  =  r.    Then  x*dx  =  \zdz,  and  the  new  integral  in  z  is 

Replacing  z  by  its  value,  we  have 

fx5(l  +  'J  a  ")^/x  =  4',  (1  +  2  x3)§  (3  z»-l)  +  C. 

Ex.  4.    Find  the  value  of    f(g  +  2)    ~  (*  +  2)    ,/.,, 
J         (a:  +  2)*  +  1 

Here  we  assume  a:  +  2  =  2*.   Then  x  =  z*—  2,  and  dx  =  4  z*dz.   On  substi- 
tution the  integral  becomes 

*/7Tr''-'  =  ,/(-',-2'-s  +  2-'2-2--  +  2-rfT)"-- 

=  I  [  l  -'■  -  i  --*  +  I  «*  -  *  +  2  *  -  2  !<*(*  +  !)]+<  '■ 

Replacing  z  by  its  value  (x  +  2)*,  we  have 

rfr  +  2)*-(x  +  2)tf/r  =  4  t_  8  t_      r+      i 

J  (,+2)Ul  5 

+  8  (x  +  2 ) '  -  s  1,  ,g  [(,■  +  2)*  + 1]  +  C. 


III.  Integrand  involving  y/a*—a?.    Let  a;  =  a  sin  2. 

Ex.  5.    Find  the  value  of    f  \  <r  -  x2./x. 
Let  x  =  a  sin  z.    Then  dx  =  a  cos  z  dz  and  V  a2  —  x2  =  a  cos  z. 
Therefore   JV«2-xVx  =  a2  f  cos2zaz  =  £  a2  f(  1  +  cos  2  z)  rfz 
=  J«a(z  +  |  sin  2  c)  +  C 


But  z  =  sin-1-  •  and  sin  2  z  =  2  sin  z  cos  z  =  2—  v  a2  —  x2. 

r/  a- 

Finally,  by  substitution,  we  have 

fVa2  -  x*dx  =  *  (x  V«2  -  x2  +  a2  sin-1  -)  +  C. 


240  INTEGRATION 


IV.  Integrand  involving  V a?  +  a2.    Let  x  =  a  tan  z. 

Ex.  6.    Find  the  value  of  J  - 


dx 


(2*  +  erf 

Let  x  =  a  tan  z.   Then  dx  =  a  sec2 zdz  and  Vj:2  +  «'2  =  a  sec  2. 

Therefore        |  =  —  /  —  =  —   /  cos  zJz  =  —  sinz  +  C. 

J(x2  +  rr)2       «"^sec2      a2  J  a2 

But  tan  s  =  —  >  whence  sin  z  =  — ===  >  so  that,  by  substitution, 
«  Vx3  +  a2 


./' 


dr 


(x2  +  rt'-)2      a2Va;2  +  a2 
If  we  try  to  find  the  value  of  f  Vx2  +  a2dx  by  the  substitution  x  =  a  tan  z, 
we  meet  the  integral  a2  C  sec3  zdz,  which  is  not  readily  found.    Accordingly 
for  a  better  method  see  Ex.  6,  §  119. 


V.  Integrand  involving  Vx2—  a2.    Let  x  =  a 
Ex.  7.    Find  the  value  of   ix3Vx2  —  trdx. 


Let  x  =  a  sec  z.    Then  dx  =  a  sec  z  tan  z  r/z,  and  Vx2  —  u2  =  a  tan  z. 
Therefore    J  x3  Vx2  —  a2dx  =  a5  (  tan2z  sec4  zdz 

=  a5  I  (tan2z  +  tan4  z)  sec2  zdz 

=  rt5(i  tan3z  +  l  tan5z)  +  C. 

x  Va-2  —  a2 

But  sec  z  =  -  ,  whence  tan  z  = >  so  that,  by  substitution,  we  have 

a  a 

fx3Vx2-a2dx  =  TV  V(.r2-«2)3(2  o2  +  3r)  +  C. 

We  might  have  written  this  integral  in  the  form    C x2V x2  —  a2  (x  dx) 

and  let  z2  =  x2  —  a2. 

VI.  Integrand     of    the     form    ,     „      ,   =  •     Let 

S  J  J  (Ax  +  B)Vax*  +  bx  +  c 

Ax  +  B  =  — 

z 

C                     dx 
Ex.  8.    Find  the  value  of   I  


:  +  l 

1 

Then  a; 

J  (2x 

2\z 

+  iy 

y$x2 

dx  = 

+  8a; 

1 

2  z' 

+  3 

Let  2  a 

and  V5  x2 

+  83 

:  +  3 

f.v* 

+  6; 

:  +  5. 

SUBSTITUTION  241 

Therefore 

/• dx ___   r  dz r  dz 

J  (2  x  +  1)  Vo  a.3  +  8  x  +  3  J  Vz2  +  t>  z  +  5         J  V(z  +  3)2  -  4 

=  -  logO  +  3  +  VY2  +  tiz  +  5)  +  C. 

But  2  = >  and  hence 

2x  +  l 


log (z  +  3  +  Vz2+6z  +  5)  =  -  log 


6  x  +  I  +  2  V.j  -r-  +  8  ,t-  +  3 
2  x  +  1 


Or  +  1 

log *+  -  log  2. 

3  a: +  2  +V5a;2  +  8x  +  3 


Therefore 


/ /*  =  leg 2J+1  +  C, 

J  (2  x-  +  1)  V5  x-2  +  8  x  +  3  3  z  +  2  +  \  5  x-  +  8  x  +  3 

log  2  having  been  made  a  part  of  the  constant  of  integration. 


118.  The  evaluation  of  the  definite  integral   I  f(x)dx  may 


*£«' 


be  performed  in  two  ways,  if  the  value  of  the  indefinite  integral 
is  found  by  substitution. 

One  method  is  to  find  the  indefinite  integral  as  in  the  pre- 
vious article  and  then  substitute  the  limits. 


Ex 


.  1.   Find  fVa2  — x«  <fe. 

By  Ex.  5,  §  117, 

fVa*  - x*dx  =  \(x  \V  -  x-  +  a2 sin-1')  +  C. 

Therefore       I     Va-  —  x'-dx  =  I  —  ( x  \  <r  —  x2  +  a'sin-1-  J 

=  _^(,-_„-  +  ,rsm-i_) 

-  -(()  Va2  -  0  +  ^sin-1-) 


A  better  method  is  to  replace  the  limits  of  /  f(x)dx  by  the 
corresponding  values  of  the  variable  substituted.  To  see  this,  sup- 
pose that  in  j  f(x)  dx  the  variable  x  is  replaced  by  a  function 
of  a  new  variable  z,  such  that  when  x  varies  continuously  from 


242  INTEGRATION 

a  to  b,  z  varies  continuously  from  eQ  to  zx.     Let  the  work  of 
finding  the  indefinite  integral  be  indicated  as  follows : 

ff(x)  dx  =  J4>  («)  dz  =  <!><»  =  F(x), 

where  F(x)  is  obtained  by  replacing  z  in  <I>  (2)  by  its  value  in 
terms  of  .r.    Then 

But  F(h)-F(a)=    C  f(x)dx, 

and  *W-^(«o)=  f^OOcb. 

Hence  f  /(.r)  cfo  =    f " '<£  (.2)  efe. 

Applying  this  method  to  the  example  just  solved,  we  have  by  Ex.  5,  §  117, 
I  v«s  —  x'2dx  =  r/2  I  cos2~<7~ 

=  $a?(z  +  £  sin  2  z)  +  C, 
where  a;  =  a  sin  z.    When  x  =  0,  z  =  0,  and  when  x  —  a,  z  —  —,  so  that  z 
as  a;  varies  from  0  to  a. 

Therefore  (    Va'2  —  x2dx  —  a2  f'2  cos2  zdz 

Jo  Jo 

-[*"(■  + t-«')I 

_  7Trt2 

In  making  the  substitution  care  should  be  taken  that  to 
each  value  of  x  between  a  and  b  corresponds  one  and  only 
one  value  of  z  between  zQ  and  z%,  and  conversely.  Failure  to 
do  this  may  lead  to  error. 


Ex.  2. 


Consider  j  2  cos  <£  dcf>,  which  by  direct  integration  is  equal  to 


=F  dx 
Let  us  place  cos  (f>  =  x,  whence  <f>  =  cos-1  a;  and  dd>  =  —  where 

Vl  -  x2 

the  sign  depends  upon  the  quadrant  in  which  <f>  is  found.    We  cannot, 


INTEGRATION  BY  PARTS  243 

therefore,  make  this  substitution  in  J  2  cos  <£  d<f>,  since  <f>  lies  in  two  differ- 
ent quadrants  ;  but  we  may  write         ~  2 

f  2  cos  <j>  d<f>  =  f    cos  <f>  d<f>  +  J  *  cos  <f>  d<f>, 

and  in  the  first  of  the  integrals  on  the  right-hand  side  of  this  equation 

place  <b  =  cos-1  a:,  dd>  =      ,  and  in  the  second  d>  =  cos-1  x,  dd>  =  —  • 

Then  V1--  VI -x- 

f^cos  *  a*  =  r-^=  -  r°_^^  =  o  r  1_^=  =  2. 

J_Z  ^      J  0  Vl  -  X2       J  !  Vl  -  X2  J  0  Vl  - ., - 

2 

119.  Integration  by  parts.  Another  method  of  importance  in 
the  reduction  of  a  given  integral  to  a  known  type  is  thai  of 
integration  hy  parte,  the  formula  for  which  is  derived  from  the 
formula  for  the  differential  of  a  product, 

c?(w)  =  udv  +  vdu. 
From  this  formula  we  derive  directly  that 
uv      I  udv  +  I  vdu, 
which  is  usually  written  in  the  form 

/  udv  =  uv  —  I  vdu. 

In  the  use  of  this  formula  the  aim  is  evidently  to  make  the 
original  integration  depend  upon  the  evaluation  of  a  simpler 
integral. 

Ex.  1.    Find  the  value  of  f  xe*dx. 

If  we  let  x  =  u  and  tFdx  =  <lr,  we  have  <lu  —  dx  and  v  =  eF. 
Substituting  in  our  formula,  we  have 

J  xexdx  —  are*  —  J  >  '  dx 

=  .rr>-  -  &  +  (  ' 
=  (X_l)fix  +  C. 

It  is  evident  that  in  selecting  the  expression  for  dr  it  is  desirable,  if 
possible,  to  choose  an  expression  that  is  easily  integrated. 


244  INTEGRATION 

Ex.  2.    Find  the  value  of  J  sin-1x^x. 

dx 
Here  we  may  let  sin-1  a:  =  u  and  dx  —  dv,  whence  du  =       '         and  v  =  x. 

Substituting  in  our  formula,  we  have  VI  —  a; 


/C     x  c 
sm~1xdx  =  x  sin_1x  —  I  — = 


xdx 


VI  -  x2 
=  a:  sin-1.*:  +Vl-  x2  +  C, 
the  last  integral  being  evaluated  by  (1),  §  116. 

Ex.  3.   Find  the  value  of  /  xcosPxdx. 
Since  cos2x  =  i(l  +  cos  2  x),  we  have 

/l   r  x2      1   r 

x  cos2 xdx  =  -  J  (x  +  x  cos  2x)dx  =  —  +  -  J  x  cos  2  xdx. 

Letting  x  =  u  and  cos  2  xdx  =  dv,  we  have  du  =  dx  and  v  —  1  sin  2  x. 
Therefore  /  x  cos  2  xdx  —  —  sin  2  x  —  -   j  sin  2  xdx 

=  -  sin  2 x  +  -  cos  2x  +  C. 
2  4 

/x2      1  /x  1  \ 

x  cos2xf/x  = 1 —  ( -  sin  2  x  +  -  cos  2  x)+  C 
4       2  \2  4  / 

=  \  (2  x2  +  2  x  sin  2  x  +  cos  2  x)  +  C. 

Sometimes  an  integral  may  be  evaluated  by  successive  inte- 
gration by  parts. 

Ex.  4.    Find  the  value  of   /  x2exdx. 

Here  we  will  let  x2  =  u  and  ex  dx  =  dv.    Then  du  =  2  x  dx  and  v  =  ex. 

Therefore  fx2exdx  =  x2ex  —  2  Cxexdx. 

The  integral  /  xexdx  may  be  evaluated  by  integration  by  parts  (see  Ex.  1), 
so  that  finally 

Cx2exdx  =  x2ex  -  2  (x  -  1)  ex  +  C  =  ex  (x2  -  2  x  +  2)  +  C. 

Ex.  5.    Find  the  value  of  J  e000  sin  bx  dx. 
Letting  sin  bx  =  u  and  e°~x  dx  —  dv,  we  have 

/eax  sin  bxdx  =  -  eax  sin  bx  — -   I  eax  cos  &x  rZx. 
a  a  J 


INTEGRATION  BY  PARTS  245 

In  the  integral  I  e"*  cos  bxdx  we  let  cos  bx  =  u  and  eaxdx  =  dv,  and  have 

/eax  cos  bxdx  —  —  e"*  cos  6x  -\ —   |  eaj"  sin  bx dx. 

Substituting  this  value  above,  we  have 

/e™  sin  bxdx  =  -eax  sin  bx (-  e"*  cos  bx  +  -   I  eai  sin  bxdx). 
a                       a  \a  a  J  / 

Now  bringing  to  the  left-hand  member  of  the  equation  all  the  terms 
containing  the  integral,  we  have 

(1  +  — )   I  eax  sin  bx  dx  =  -  e**  sin  bx e"  cos  bx, 
a- J  J                             a                       «'- 


whence 


/•     .     ,        eax  (a  sin  bx  —  h  C<  is  bx ) 
e^  sin  bx  dx  = * : ' 
a2  +  U1 


x  dx 


J       Ex.  6.    Find  the  value  of  f  Vx2  +  a?dx. 

Placing   Vx2  +  a2  =  u   and  dx  =  dr,  whence   du  =  — =      =  and    v  =  x, 
we  have  A  ■'"  +  "" 

fVx*  +  a*dx  =  x  Vx*  +  a*  -  f    xVx     .  (1 ) 

J  J  Vx2  +  «2 

Since  x2  =  (x2  +  a2)  —  a2,  the  second  Integra]  of  ( 1  )  may  be  written  as 
r(x2  +  a2)dx  _    „  r      dx 
J     Vx2  +  a2  J  Vx2  +  «'-' 

which  equals  /  vr  +  crdx  —  a*  j  —  • 

J  J  a  ./-  +  a2 

Evaluating  this  last  integral  and  substituting  in  (1),  we  have 

f  Vx2  +  u2dx  =  x  Vx2  +  a2  -  f  Vx2  +  crdx  +  a2  log  (x  +  Vx2  +  a2), 

whence       fVx2  +  irdx  =  },  [x  Vx2  +  a2  +  a2  log  (x  +  Vx2  +  a2)]. 

120.  If  the  value  of  the  indefinite  integral   if(ai)dx  is  found 
by    integration   by    parts,    the    value    of    the    definite    integral 

X/(V)  dx  may  be  found  by  substituting  the  limits  a  and  b,  in 
the  usual  manner,  in  the  indefinite  integral. 


246  INTEGRATION 

Ex.    Find  the  value  of  C  2  x2  sin  x  dx. 

To  find  the  value  of  the  indefinite  integral,  let  x2  =  u  and  sinxdx  —  dv. 
Then  j  x2  sin  x  dx  =  —  x2  cos  x  +  2  I  x  cosxdx. 

In  /  x  cos  a; (for,  let  x  =  w  and  cos  xdx  =  dv. 

Then  I  x  cos  a;  Jx  =  x  sin  x  —  /  sin  x  dx 

=  x  sin  x  +  cos  x. 
Finally,  we  have 

j  x2 sin  xdx  =  —  x2  cos  x  +  2x  sin  a;  +  2  cos  a;  +  C. 

Hence  f  2  x2  sin  x  c/x  =    —  x2  cos  x  +  2  x  sin  x  +  2  cos  x 

=  tt-2. 

The  better  method,  however,  is  as  follows :  6 

lif(x)dx  is  denoted  by  udv,  the  definite  integral   j  f(x)dx 

Xb  J  a 

udv,  where  it  is  understood  that  a  and  b 

are  the  values  of  the  independent  variable.    Then 
I    udv  =  [uv~\  —  /    vdu. 

To  prove  this,  note  that  it  follows  at  once  from  the  equation 

/*>b  s%b  />6  s*b 

[uv~\  =  j    d(u,v*)  =  j    (udv  +  vdu)  =  j    udv  +   I    wc?m. 

Applying  this  method  to  the  problem  just  solved,  we  have 
C  2"  x2  sin  x  rfx  =  I  —  x2  cos  x     +  2  C  2  x  cos  x  rfx 

=  2   r  2  x  cos  x  dx 

Jo 


2  sin  x  rfx 


o 


PARTIAL  FRACTIONS  247 

121.  Integration  by  partial  fractions.  A  rational  fraction  is  a 
fraction  in  which  both  the  numerator  and  the  denominator  are 
polynomials-  If  the  degree  of  the  numerator  is  equal  to,  or 
greater  than,  the  degree  of  the  denominator,  we  may,  by  actual 
division,  separate  the  fraction  into  an  integral  expression  and  a 
fraction  in  which  the  degree  of  the  numerator  is  less  than  the 
degree  of  the  denominator. 

For  example,  by  actual  division, 

2*5-.r4  +  .r3+3./--3i;.r  +  :)(;      n_  ,r»+3^_4J.+  20 

"•"  ..4       1/;  '     {    ) 


xA- 10  z*-16 

It  is  evident,  then,  that  we  need  to  study  the  integration  of 
only  those  fractions  in  which  the  degree  of  the  numerator  is 
less  than  the  degree  of  the  denominator. 

If  the  denominator  of  such  a  fraction  is  of  the  first  degree 
or  the  second  degree,  the  integration  may  be  performed  by 
formulas  (2),  (14),  (18),  §116,  as  in  Ex.3,  §  114. 

If  the  denominator  is  of  higher  degree  than  the  second,  we  can 
separate  the  fraction  into  partial  fraction*  the  sum  of  which  will 
equal  the  given  fraction. 

For  example, 

a3+3J.3-4a:4-20_     1  1         x-1 

.r'_16  ~x-2     x+2+a?+4' 

as  the  reader  can  easily  verify. 

The  three  fractious  on  the  right-hand  side  of  (2)  are  the 
partial  fractions  of  the  fraction  on  the  left-hand  side  of  (2). 
It  is  to  be  noted  that  their  denominators  are  the  rational 
factors  of  the  denominator  of  the  fraction  of  the  left-hand 
side  of  (2). 

Substituting  in  (1),  we  have 

2  :r5_  .r-»+  a*+  ;j  ,r-  30  x  +  30 
a;4- 10 


248  INTEGRATION 

r  2  x5-  x4  +  xa+  3  x2-  30  x  +  36  , 
Hence    J  -4 — 3— ax 


dx 


^_:r+log(r_2)_iog(:r+2)  +  ilog(^+4)-itan-1^ 


,  .      (.r-2)V*2+4      1,     _xx 
=  x2-x  +  \og^ -^ 2tM      2' 

The  separation  of  a  fraction  into  partial  fractions,  as  in  (2), 
is  evidently  a  great  aid  in  integration.  We  shall  illustrate  this 
process  in  the  following  examples : 

Ex.  1.    hnd  the  value  of   I —  dx. 

J  (x  +  3)  (x2  —  4) 

The  factors  of  the  denominator  are  x  +  3,  x  —  2,  and  x  +  2.  We  assume 
x2  +  11  x  + 14    _     ^  £  C  a 

(x  +  3)  (x2  -  4)      x  +  3       x  -  2      x  +  2  '  ; 

where  A ,  B,  and  C  are  constants  to  be  determined. 

Clearing  (1)  of  fractions  by  multiplying  by  (x  +  3)  (x2  —  4),  we  have 

x2+llx  +  14  =  .4(x-2)(x  +  2)+i3(x+3)(x  +  2)  +  C,(x  +  3)(x-2),  (2) 

or    x2  +11  x  +  14  =  (A+B  +  C)za  +  (55  +  C)x  +  (-4.4  +6  B  -  6  C).  (3) 

Since  .4,  B,  and  C  are  to  be  determined  so  that  the  right-hand  member 
of  (3)  shall  be  identical  with  the  left-hand  member,  the  coefficients  of  like 
powers  of  x  on  the  two  sides  of  the  equation  must  be  equal. 

Therefore,  equating  the  coefficients  of  like  powers  of  x  in  (3),  we  obtain 
the  equations  .   ,   d  ,  ^_i 

5B+  C  =  ll, 
-  4  A  +  6  B  -  6  C  =  14, 

whence  we  find  ,4  =  - 2,  B  =  2,  C  =  l. 

Substituting  these  values  in  (1),  we  have 

x2  + 11  x  + 14  2  2       ,       1 


(x  +  3)(x2-4)  x  +  3      x-2      x  +  2 

.    r  x2  +  llx+14     ,  r2dx    ,    r2dx    ,    r    dx 

andi(3;  +  3)(x2-4)^="Jxn  +  i.^2  +  JxT2 

=  -2  log(x  +  3)  +  2  log(x  -  2)  +  log(x  +  2)  +  C 
=        (X  +  2H.-2)2      c> 
b         (x  +  3)2 


PARTIAL  FRACTIONS  249 

Ex.2.    Find  the  value  of  f4  ^  +  x  +  l  dx. 

J        x3  —  1 

The  real  factors  of  x3  —  1  are  x  —  1  and  x2  +  x  + 1.    Hence  we  assume 
4  x2  +  x  +  1  =     .4  Bx  +  C 

x3  -  1  a;  - 1      x2  +  x  +  1 " 

Clearing  of  fractions,  we  have 

4  x2  +  x  +  1  =  A  (x2  +  x  +  1)  +  (Bx  +  C)  (x  -  1) 

=  (.4  +  B) x2  +  (.4  -  B  +  C)x  +  (-4  -  C).  (2) 

Equating  coefficients  of  like  powers  of  x  in  (2),  we  obtain  the  equations 
A  +  B  =  4, 
A  -  B  +  C  =  1, 

A-C  =  l, 

whence  A  -  2,  B  =  2,  C  =  1. 

„                   4x2  +  x  +  4          2  2x  +  l 

lience = + 


/''  r-i/r 
; 
(x  +  2)2(x-2 


B8  -  1  x  -  1.        X2  +  X  +  1 

and  f4*2  +  *  +  1,/x=f^l  +f(2*  +  l)rf* 

J  x3  -  1  J  x  -  1       J      X"  +  X  +  1 

=  2  log  (x  -  1)  +  log  (x2  +  x  +  1)  +  C 
=  log  [(x  -  1)2(X2  +  X  +  1)]  +  C. 
•2  x*dx  _ 

Here  we  assume 

(x  +  2)2(x-2)       (x  +  2)2      x  +  2      x-2  w 

Clearing  of  fractions,  we  have 

2  x2  =  .4  (x  -  2)  +  B  (x2  -  4)  +  C(x  +  2)2 

=  (B  +  C)x*  +  (A  +  \C)x  +  (-2A  -4B+4C).  (2) 

Equating  the  coefficients   of   like  powers  of  x  in  (2),  we  obtain  the 
equations  B  +  C  —  2 

A  +  4  C  =  0, 
-  2  /I  -  4  B  +  4  C  =  0, 
whence  y4  =-  2,  5  =  f,  C  =  .1. 

2x2  =  2  g  &     ^ 

(x  +  2)2  (x  -  2)  (x  +  2)2      x  +  2      x  -  2  ' 

J  (x  +  2)2(x  -  2)         J  (x  +  2)2     J  x  +  2      J  x  -  2 

=  x~T^  +  I l0S  (''  +  2)  +  ^l0g  <*  "  2)  +  ° 


— =—  +  log  V(x  +  2)3(x  -  2)  +  C. 
x  +  J 


250  INTEGRATION 

Ex.4.    Find  the  value  of  f^  X+  +  (  *  ~  ^  dx. 

Now  (x  +  1)  (x8  +  1)  =  (x  +  l)a(ar2  —  x  +  1),  and  we  assume 
3x3  +  3x-6  J  .       /;      |     Cx+  I> 


(x  +  1)  (a;8  +  1)       (x  +  If      x  +  1      x2  -  x  + 1 
Clearing  (1)  of  fractions,  we  have 
3  x3  +  3  x  -  6  =  A  (,i-2  -  x  +  1)  +  B (x8  +  1)  +  (C x  +  D)  (x  +  l)2 

=  (B+  C)xs  +  (A  +  2  C  +  Z>)x2  +  (-  .1  +  C+  2D)x 

+  (A  +  B  +  D).  (2) 

Equating  coefficients  of  like  powers  of  x  in  (2),  we  obtain  the  equations 
B  +  C  =  3, 
A  +  2C  +  D  =  0, 
-A+C  +  2D  =  3, 
A  +  B  +  D  =  -  6, 

whence  .4  =  -  4,  5  =  0,  C  =  3,  7)  =  -  2. 
Substituting  these  values  in  (1),  we  have 

3  xs  +  3  x  -  6  -  4  3  x  -  2 


(x  +  1)  (x3  +  1)       (x  +  4)2      x2  -  x  +  1 

f3.*3  +  3x-<!^=  f-4^  +  f   3*~2    dx 
J  (x  +  1)  (x3  +1)  J  (.'■  +  l)2      J  x1  -  X  +  1 


4.  3  1  2  r  —  1 

=  -i-  +  -  log  (x2  -  x  +  1) ==  tan-1     V     +  C, 

x  +  1      2  V3  V3 

the  last  integral  being  evaluated  as  in  Ex.  3,  §  114. 

We  notice  in  the  solution  of  the  above  examples  the  follow- 
ing points : 

1.  The  denominator  is  factored  into  linear  or  quadratic  factors, 
or  integral  powers  of  such  factors. 

2.  As  many  partial  fractions  are  assumed  as  there  are  factors  in 
the  denominator. 

3.  Corresponding   to    any   single   linear  factor,   as   ax  -+-  b,   one 

fraction  of  the  form is  assumed,  and  corresponding  to  the 

ax  +  o  ^ 


square  of  any  linear  factor,  as  (ax  +  b~)2,  two  fractions  - —      ,.  2 

7? 

H are  assumed,  the  numerator  over  the  square  of  the  factor 

ax  +  b 

being  of  the  same  type  as  that  over  the  first  power  of  the  factor. 


PARTIAL  FRACTIONS  251 

4.  Corresponding  to  any  single  quadratic  factor,  as  ax*  +  bx+c, 

Ax  +  T> 

one  fraction  of  the  form  — 5 is  assumed. 

ax  -f-  ox  +  c 

5.  The  numerators  assumed  are  determined  and  the  integration 
of  the  partial  fractions  is  completed. 

If  (ax  +  by  in  3  is  replaced  by  (ax  +  6)B,  and  the  correspond- 
ing n  fractions  are  assumed  to  be 

A    + * +...+-*-, 

(ax  +  of1      (ax  -\-  by   1  ax  +  b 

and  if  ax2  +  bx  +  c  in  4  is  replaced  by  (ax2  +  bx  +  c)n,  the  cor- 
responding n  fractions  assumed  being 

Ax  +  B  Cx  +  D  Px  +  Q 

(ax2  +  bx  +  c)n      (ax2  +  bx  +  c)n  _1  <i.r  +  6a;  -f-  c ' 

the  above  becomes  a  working  rule  for  the  integration  of  all 
rational  fractions  in  which  the  degree  of  the  numerator  is  less 
than  the  degree  of  the  denominator;  but  the  proof  of  the  pos- 
sibility of  assuming  the  partial  fractions  in  the  form  noted 
above  is  omitted. 

To  make  the  work  of  this  article  complete  we  must  discuss 

the  integral   / ^-J— - —  — -dx,  where  n  is  any  integer  greater 

J    (''■>"  +  OX  +  CJ 

than  unity. 

Since  d  (ax2  +  bx  +  (?)  =  (2  ax  +  i)  dx,   we   may,  as  in  Ex.  3, 

§114,  let  Ax  +  B  =  -£-(2ax  +  b')  +  B-  —  ,  and  obtain  the 
equation 

C     Ax+B      dx=J±  rd(ax*+bx+c)     / _  Ab\  C        dx         _ 

J  (ax2+bx+cj    3      -2aJ  (<tx2  +  bx+cy     \        2a/J (ax*+bx+cy' 

Proceeding  as  in  Ex.  2,  §  114,  we  may  put  the  last  integral  in 

the  form  /  — - — -— ,  which  may  be  reduced  to  the  integral  /  — — - 
J  (u*+a2y  J  &     J  u2+a2 

by  successive  applications  of  the  formula 

r   <**    _     1     r    u     \(2n  s-)  f - 1. 

J  (u2+a2y      2<>-iyr[(w2+«2)»-1     v  JJ  (w2+a2)B_1J 

This  is  a  special  case  of  (4),  §  122. 


252  INTEGRATION 

122.  Reduction  formulas.  The  methods  of  integration  derived 
in  this  chapter  are  sufficient  for  the  solution  of  most  of  the 
problems  which  occur  in  practice.  If  the  reader  should  meet 
any  integrals  which  cannot  be  evaluated  by  these  methods,  he 
should  refer  to  a  table  of  integrals,  in  which  the  integrals  have 
been  either  completely  evaluated  or  expressed  in  terms  of  simpler 
integrals.  Some  of  this  latter  type  of  integrals,  known  as  reduc- 
tion formulas,  have  been  tabulated  below  for  convenience. 

= * -fr r zrrr  I  %     n(a  +  bxnydx,     (1) 

(np  +  m+V)b         (np  +  m+l)bj  J  v  J 


^far(a  +  bx»y-idx,  (2) 


Cxm(a  +  bxnydx 
_xm  +  1(a  +  bxn] 

np  +  m  +  1  np  +  ni 

C  xm  (a  +  bxnyp  dx 

^Xa  +  tey«_(np  +  n  +  m+V)b  f^+.(fl+tey^    (3) 
(m+V)a  (m  +  l)a       J  K  T      J      '    ^  J 

fxm(a  +  bx'l)pdx 

=_ary+bz»y~     y  +  n+m+l  C^a+^y+idx.     (4) 
n(jt?+l)a  n(p+l)a  J       K  J  K  J 

j  sinma;  cos"  xdx 


/' 


sin"1  #  cos"x^» 


/sinm^  cosn_2a:  cfe,  (5) 

+  w  4-  2  /* 

—  /  sinm:zcos"  +  22;efo,  (6) 

n+1     J  K  J 


x      m 


71  +  1 

j  sinm:z  cosnxdx 

-\ — j  sinm~2xcosnxdx,  (7) 

m  +  n  J 


sinm_1:r  cos"+1 


f< 


m  +  n 
Sin""  a;  cos" xdx 

sinm+1a;  cos"+1x      m 


m  + 1  m 


+  n  +  2   r  . 
; —  I  S11 


lxcosnxdx.  (8) 


PROBLEMS  253 

These  formulas  do  not  always  hold.  For  example,  (1)  and 
(2)  fail  if  np  +  m  +  1  =  0,  (3)  fails  if  m  +  1=0,  etc.  In  these 
cases,  however,  it  is  not  necessary  to  use  these  formulas,  as  the 
integration  may  be  performed  by  elementary  methods. 

There  are  also  integrals  which  cannot  be  expressed  in  terms 

of  the  elementary  functions.    For  example,   /  — 

J  V(l-z*)(l-*V) 

cannot  be  so  expressed;    and,   in  fact,   this   integral  defines  a 

function  of  x  of  an  entirely  new  kind. 

PROBLEMS 

Find  the  values  of  the  following  integrals  : 

Jl.    f  (4  x3  +  3r  +  4x-3)  dx.        12.    f-x  +  e_"x  dx. 
2.  f(a*  -  x*  +  i  -  i)«fa  13.  f(2  +  Sx)*dx. 

J  \  <x*J  J  1+2S-3*8 

4.     ft+^tl^  15.     fj» 

J  Vx  J   «  + 


3  xdx 

b  sin; 


6.  r&+ffifc  „.  r  ('+*>'*  . 

J        x"  J   Vl+3a;+a:8 

7.    /  V2  +  e^tfe.  18.    /  ^t— jr-ete. 

J  J   <-c  —  cos  2x 

Pe**dx  C dx 

'Je**  +  2'  19J  (x  -a)  [log  (x  -«)]«' 

9.     /  — r-  20.     /    -^<£c. 

J  xlogx2  J       X 

10-/(i  +  ^.-/  n-fp  +  f)*.*.- 

n.  r7i±^5<fa.  22.  f(J=B-A± 

J   (x  -  cos  a-)2  J  \  \«  +  x       \|«  - 


dx. 
x, 


254  INTEGKATION 


23. 


24. 


/, 7^ .  40.   /  (am|-j-cosf  )da;. 
Vaa  -  «2  log  (as  +  Vx2  -  d2 )  J  \       3  3/ 

25.  J  cos3x  sinxdx.  41.    /  tan3  -  da;. 

26.  fain8(2a;+l)cos(2a;+l)da:.  42.    fsec4 (3 a;  +  2) da;. 

//•       3  T         3  x 

(sec  «x  -4-  tan  ax)  sec  aa;da;.  43.    J  tan  —  sec3  —  da;. 

/  (esc  a;  etna;  +  esc  a-)'2 da;.  44.    J  sec4  2xVtan  2xdx. 

/  cos2  2  x  sin3  2  a;  da-.  45 .    J  esc6  ^  da?. 

/  sinf  3  x  cos5  3  a- dx.  46.    J  ctn5 (x  4-  2)  da;. 

/.   „x  r        2x         2x 

sms  g  "x-  47 .    I  ctn2  —  esc4  —  dx. 

/cos3  4  x  /* 

,  -dx.  48.    I  tan4  ax  da;. 

Vsin3  4  a;  J 


28 


29 


30, 


?,2 


esc  ^.r 


/cos  2  x  _  A 
da-.  49.   I  - 
c6s  a?                                                        j    1 

34.  I  sin  a;  sin  2  a- da;.  50.    /  tan3  ^  "\J  sec  |  da;. 

35.  I  (tan  ax  4- ctn«x)2dx.  51.    J  sin2(3x +l)dx. 

fesc3  3  x  -  ctn3  3  x  7  /\  .    _  _    N0  . 

37.'/ - ; — 7: — dx.  53.    I  (sin  2x  —  cos  2xrdx. 

J    esc  3  x  —  ctn  3  x  J 

38.    I  ( tan2 -  —  ctn2 -  J  dx.  54.  I  sin2  3x  cos2  3xdx. 


PROBLEMS  255 

55.    I  cos4'-r/./\  71.    |  (1—  cos4x)^ox. 


56. 


Jo  -  9  x2 


/  cos4|ox.  71.    /  ( 

I  sin4x  cos2xax.  72.    I  — - = 

57.    /  Vcos2x  +1  sin  2xdx.  73.    I  • 

68.   /7^  _  £2il^\  fc  74.   f        *         ■ 
J   y  sin  a;          cos  x  /  J  /V4  ar  —  9 

f    cos2.Wx  75      f  '7'" 

59-Jcosx  +  sinx'  'J  4.,2  +  4x+10 

60.  I  sin  ax  sinbxdx.  (a^b)  76.    /  .,  r<   i    5  ,.   .   4 ' 

61.  I  cos  ax  cos  focrfx.  (a  =£  A)  77.    I       _- __ -. 

62.j"sin(2x  +  3)cos(2x-3),/r.  78.  J  ^_^_==. 

63.    I  sin x  sin  2 x  sin  3. '•'/.'•.  79.    I  - 


./■  —  ./- 


67 


r       sec  2  x  ox  80     f ^ 

"  J  sec  2  x  —  tan  2  x  '  '  J  (x  +  1)  Vx2  +  2  x 

_  r  |cscx-ctn,f/),  81    f4±;  /,, 

J     >  esc  x  +  ctn  x  J  •'    +  J 

J    1  —  COS  X  J   • » •'    +  - 

/ci.s.i-f/x  83^  r  x  —  2 

1  +  COS  X  J    Vl  —  X2 


2^' 


7 84-    I  -7==  d». 

sec  aa  J   y,/  _  ,,■ 

/. f  sinxc/x 

Vl-cosxr7x.  85.  j   1+cos2: 

J   Vl+cos2x  J    V4x 


70 


il.r 
IS 

<l.r 


+  9 


256  .  INTEGRATION 


88, 


95 


96 


97 


99. 


3  —  3 


dx 

+  sec2  a: 


/dx  r  dx 

9«3-2*  ^  J,2 +  2*  tan* 

r      dx  r 

I       /  105.     | 

J   V4-ca  +  6a;  J  e2x  + 


/^&'  106-/vS 


2  ex  sin  a  + 1 


r'  tan  a  —  e2 


91-  /  ~^~r^ — rr-  107-  I  — ^=^- 

92.    f  ^  108.    fi^ 

J    V3^-2x  +  3  J  x-V^  +  «2 

93-/2l-t-^  109.  J 


dx. 

+ 


4  —  5  COS  ; 


94.  f       dx  110.  r     ^ 

J    V2x2  +  4x  —  7  J  4  +  3sinx 

r  (5x-S)dx  r        dx 

J  x2  + 63+12"  in'  J  6  +  2cos2a;' 

Jx*  +  x-S.  112'JiT^ 
f(2x+10)dx  r 

J  2aJa  +  5»+i"  113-J  ; 


sin  4  a; 
dx 


cos  cc  —  2  sin  a; 
)2<Zx. 


r(6x  +  20)dx  r 

J  6x2  +  73-3'  114'J  (eS  +  e  * 

J  2  a3 +  6* +  9*  115- J  (^  +  «0*b. 


r     (a?-2)rfa;  /* 

J  3^  +  23  +  3'  116'J 


(c1"  +  e~^)xdx. 


id.  r  (*+w*=.        117.  rv^. 

J    V3  +  2x-x2  J  *2 

102.     /        /  ;=-  118.     /   e«n    Ja 


<7.r 


Vi-»2 


PROBLEMS  257 


122 


123 


124 


126 


127 


.    fe*+Wdx.  135.    f    *''d*   , 
J  J  (•'•  +  3)'2 

.    fe»+™ab  +  ™dx.  136.    C    ****  . 
J  J  3  +  4  a;3 

'  f^Tidx-  137.  ^V^: 

/  -jz Tdx.  139.    / 

J   V2z-3  J         x 


4)i 
dx. 


fV*_+2+l_  fV4/-9  , 

.  r  *'<**  .  142.  r.  ^  . 

J  (aj'  +  a8)*  J  (3  +  2sa)* 

.  r-**-.  143.  r  -v-  . 

J    (a8 +  3)'  J    (-'-  +  9)5 

128.    f    ,  =•  144.    f  (a"  —  jcMefe. 

J   V(4  -  a2)8  J 

129.  r  *Va;  .  145.  r,(,-+i)?^. 

J  V(9  -  x2)5  J 

130.  r^(4 + ^j,;,.  146.  j^-^y  dx. 

131.    f   '       *  ■  147.    f— *=• 

J/V4  +  9ic2  J  a!*V2-a9 

!82.    f^^,,,,,  148.    ffit±a»fc 

J  X  J  X 

/dx  C      x3dx 

aj8V9K9-4'  U9*  J   VT+472' 


133 


258  INTEGRATION 

,  5*  166.    /  xsin*3xdx. 

152.    /  -        — .  167.    Ce2xamxdx. 

J  (rc-2)V2ar8-4a;-l  J 

153 


r  rfa  f 

/  — ,  =^-  168.    /  e*cos3aufa. 

J  (2jc-1)V16k2-12cc  +  3  J 

^=-  169.    /  Vai2— ltfa 

(2.r-3)V4*2-12a-  +  5  J  1^' 

C  dx  r    , 

/  ,  =•  170.    /  sec3a;( 

J  (.t  +  2)Vx-'2  +  2x  +  2  J 


155.  I  =r-  170.    |  sec3Z(fa. 

J   (.T  +  2)Va;2  +  2z  +  2  J 

156.  flogaxdx.  171.    f    1t'r'     (fa- 


157. 


+  4) 
dx. 
dx 


158.  /  t&n~1axdx.  173.    /  — - 
J  J  ■'■(■>■ 

159.  /%,-(.,. +  Va-2  +  «2)^.  174.     f^f^ 

160.  |  a-  sin  3 a; (fa.  175.    f — — 
fsee->2xdx.  176.  f^±< 

Ixsec-^xdx.  177.    f   *  +  40,  <fa. 


161.  I  sec-^xffa.  176.    /  r^da, 

162.  /  x  sec^Sxdx 

163.  I  x*eixdx 

164.  I  ai2cos  2x(Zx 


J  scVcfa.  178.    J 


+1) 
Id 

2  a; -8) 


PROBLEMS  259 


181.    f- ^—r  — —  dx.  /^sin 


x2  -  2  x  +  3 
x-2)(x+l)(x-S)-  ioj       /    ; 


3  I      cos  - 


183. 


184, 


185. 


186 


187 


188 


189. 


190 


192 


C  4  x-  -  3 

182.    I  — — —dx.  ^  2 

J  (x  -2)  (a»  +  2  x  +  5) 

rcos24x  , 

f8»  +  **.  195-J^T^ 

J  a;(ar-4)2 

/ardse 

/•     </.'■ 

f    da;  197.    |  7- 

Jxzi-  J  (i  +  -<-V 

/v+n,--67  198.  f-  — — -• 

.     f       *        •  200.     f— ^— r. 


/£±f^  301.  J>  +  „".,.., 

/  ^— cZa;.  202.    I  -t-= r-' 

J     1—  x  J  sura  cos^x 

191.     f*+^Eldx.  203.     f-'",- 
J   aj-Vl-a;  J  sin8x 

f  da  f       dx 

r*  204.        -r-= 

j  cos°a;  ^y  sura  cos  x 

I  7^ — T  205.    i - 

Ja(a8  +  2)*  J« 


cus5^; 


CHAPTER  XIII 

APPLICATIONS  OF  INTEGRATION 

123.  Element  of  a  definite  integral.  In  §§78  and  81,  by  means 
of  the  area  under  a  curve,  we  have  defined  the  definite  integral 
by  the  equation 

"  /(*)  dx  =  Lim  5)/(a0  Ax,  (1) 


£ 


and  have  shown  that  this  limit  may  be  evaluated  by  the  formula 
hf(x)dx==F(h)-F(a),  (2) 


I. 


where  dF(x)=f(x)dx. 

Since  any  function  f(x)  may  be  graphically  represented  by 
the  curve  y  —f(x),  formulas  (1)  and  (2)  are  perfectly  general. 
We  shall  proceed  to  give  certain  applications.  The  general 
method  of  handling  any  one  of  the  various  problems  proposed 
is  to  analyze  it  into  the  limit  of  the  sum  of  an  infinite  number 
of  terms  of  the  form  f(x)  dx.  The  expression  f(x)  dx,  as  well 
as  the  concrete  object  it  represents,  is  called  the  element  of 
the  sum. 

124.  In  finding  the  element  of  integration,  it  is  often  not 
possible  to  express  the  terms  of  the  sum  (1),  §  123,  exactly 
as  f(Xi)  Ax,  the  more  exact  expression  being  [/(#<)  +  ej  Ax, 
where  the  quantities  et  are  not  fully  determined  but  are 
known  to  approach  zero  as  a  limit  as  Ax  approaches  zero. 
It   is   consequently   of   the    highest    importance    to   show   that 

Lim  V  e.Ax  =  0,  so  that 

Lim  2  [/OO  +  eJAx=Umy/f(xi-)Ax=  Cf(x)dx. 


ELEMENT  OF  A  DEFINITE  INTEGRAL  261 

For  that  purpose,  let  7  be  a  positive  quantity  which  is  equal  to 
the  largest  numerical  value  of  any  e.  in  the  sum.    Then 

-  7  ^  e.  =  7 
and  —  27A.T  =  Se.Ar  =  ^7A.r. 

But  27AX  =  7lA.r  =  7  (b  —  a) 

and  Lim  ~2yAx  =  0  since  7  approaches  zero  as  A.r  approaches  zero. 
Hence  Lim  "Ze.Ax  =  0. 

Hence  the  quantities  et  which  may  appear  in  expressing  the  sum 
do  not  affect  the  value  of  the  integral  and  may  be  omitted. 

Quantities  such  as  Ax  and  e-,  which  approach  zero  as  a  limit, 
are  called  infinitesimal*.  Terms  such  as f(x)  A.r,  which  are  formed 
by  multiplying  Ax  by  a  finite  quantity,  not  zero,  arc  called  infini- 
tesimals of  the  same  order  as  Ax.  Quantities  such  as  e  Ax,  which 
are  the  products  of  two  infinitesimals  approaching  zero  together, 
are  called  infinitesimals  of  higher  order  than  either  infinitesimal. 

The  theorem  above  proved  may  be  restated  in  the  follow- 
ing way  : 

In  forming  the  element  of  integration  infinitesimals  of  higher 
order  than  f(x)Ax  may  be  disregarded. 

Ex.  Consider  the  area  under  a  curve  (§78).  We  have  obtained  it,  by 
means  of  rectangles,  as  i  =  „-i 

Lim  V  /(x,)Ax.  (1) 

Suppose  that  in  place  of  the  rectangles  we  used  the  trapezoids  formed 

by  drawing  the  chords  DPV  l\l'.,,  etc.  (fig.  125).  The  area  of  one  Buch 

trapezoid  is  -,   .  A     ,   ,  A   A 

1  f(Xi)Ax  +  \  A//Ar. 

But  I  Ay  is  a  quantity  which  approaches  zero  as  a  limit  when  Ax 
approaches  zero,  and  may  be  denoted  bye,--  Hence,  it'  we  used  the  trape- 
zoids, we  should  have  for  the  required  area 

Lim"j?  [/(*,)+ «*] Ax.  (2) 

"  ~  ■  « =  0 

We  see  then  directly  that  in  this  example 

i'=n-l  i  =  n-l 

Lim  J)  [/(*«)  +  */]  Ar  =  Lim  J  f(xt)  Ax. 


262 


APPLICATIONS  OF  INTEGRATION 


125.  Area  of  a  plane  curve  in  Cartesian  coordinates.  This 
problem  was  used  to  obtain  the  definition  of  a  definite  inte- 
gral, with  the  result  that   the   area  bounded  by  the  axis   of 


.r,  the  straight  lines  x  =  a  and  x—  b  (a<  6),  and  a 
of  the  curve  y  =f(x~)  which  lies  above  the  axis 
given   by   the    definite   integral 


f. 


ydx. 


portion 
of    x   is 


(i) 


It  has  also  been  noted  that  either  of  the  boundary  lines  x=  a 
or  x  =  b  may  be  replaced  by  a  point  in  which  the  curve  cuts  OX. 
Here  the  element  of  integration  ydx  represents  the  area  of  a 
rectangle  with  the  base  dx  and  the  altitude  y. 

Similarly,  the  area  bounded  by  the  axis  of  y,  the  straight  lines 
y  =  c  and  y  —  d(c  <  d),  and  a  portion  of  the  curve  x  =f(jf) 
lying  to  the  right  of  the  axis 
of  y  is  given  by  the  integral 


/ 


d 

xdy, 


(2) 


where  the  element  xdy  repre- 
sents a  rectangle  with  base  x 
and  altitude  dy. 

Areas  bounded  in  other  ways 
than  these  are  found  by  express- 
ing   the    required   area  as  the 

sum  or  the  difference  of  areas  of  the  above  type,  or  by  writing 
a  new  form  of  the  element  as  illustrated  in  Ex.  2. 

Ex.  1.    Find  the  area  of  the  ellipse  —  +  '(-  =  1. 

«2      Ir 

It  is  evident  from  the  symmetry  of  the  curve  (fig.  157)  that  one  fourth 
of  the  required  area  is  bounded  by  the  axis  of  y,  the  axis  of  x,  and  the 
curve.    Hence,  if  A  is  the  total  area  of  the  ellipse, 


4  =  4 

2  b 


f   y  dx  =  4  C    -  Vo2  -  x2 

Jo  Jo    a 

lH    n — 2  ,    2  •    1*1 

—  \x  V  a*  —  x*  +  a-  sin- 1  - 
a  L  «J 


dx 


TTClb. 


AEEA 


263 


Ex.  2.  Find  the  area  bounded  by  the  axis  of  x,  the  parabola  y-  =  4  px, 
and  the  straight  line  y  +  2x  —  ip  =  0  (fig.  15S).  The  straight  line  and 
the  parabola  intersect  at  the  point  C 
( p,  '2  p),  and  the  straight  line  intersects 
OX  at  B  (2p,  0).  The  figure  shows  that 
the  required  area  is  the  sum  of  two 
areas  OCD  and  CBD.  Hence,  if  A  is 
the  required  area, 


r 


1  px  dx  + 


f<>. 


The  area  may  also  be  found  by  con- 
sidering it  as  the  limit  of  the  sum  of 
such  rectangles  as  are  shown  in  fig.  159. 
The  height  of  each  of  these  rectangles 
is  Ay,  and  its  length  is  x„  —  xv  where  x„  is  taken  from  the  equation  of 
the  straight  line  and  x1  from  that  of  the  parabola.  The  values  of  y  range 
from  y  =  0  at  the  base  of  the  figure  to 
y  —  2p  at  the  point  C.    Hence 

-[■»-?-i£]M* 

In  the  above  examples  we  have 

replaced  y  in    J    ydx  by  its  value 

f(x)  taken   from  the   equation   of 

the  curve.    More  generally,  if  the 

equation  of  the  curve  is  in  the  parametric  form,  we  replace  both 

x  and  y  by  their  values  in  terms  of  the  independent  parameter. 

This  is  a  substitution  of  a  new  variable,  as  explained  in  §  118, 

and  the  limits  must  be  correspondingly  changed. 

Ex.  3.   Let  the  equations  of  the  ellipse  be 

x  =  a  cos  <£,         y  =  b  sin  <f>. 
Then  the  area  A  of  Ex.  1  may  be  computed  as  follows: 

J.  =  4   |     ydx— —A   j    ah  sin2  <f>t!<f>  =  4  ah  j  ~  sin2  <£  tlcf>  =  trab. 


Fig.  159 


264 


APPLICATIONS  OF  INTEGRATION 


126.  Infinite  limits  or  integrand.  If  the  curve  extends  in- 
definitely to  the  right  hand,  as  in  figs.  160-1G2,  it  is  possible 
to  consider  the  area  bounded  by  the  curve,  the  axis  of  as,  and  a 
fixed  ordinate  x  =  a,  the  figure  being  unbounded  at  the  right 
hand.    Such  an  area  is  expressed  by  the  integral 

f(x)  dx  =  Lim  F(b)  -  F(a), 

which  may  be  written  concisely  as 


jf 


f(x)  dx  =  F(oo  )  -  F(d). 


There  is  no  certainty  that  this  area  is  either  finite  or  deter- 
minate.   Where  it  is  so,  the  area  bounded  on  the  right  by  a 
movable  ordinate  approaches  a  defi-       7 
nite   limit  as  the   ordinate  recedes 
indefinitely  from  the  origin. 

[2vTT  =  ^.   (Fig.  160) 


Ex.  1 


Va 


Ex.  2 


Ex. 


I 

rs-Hr-1- 

I     sin  xdx  =  [—  cos  x^° 

—  indeterminate.      (Fig.  162) 


(Fig.  161) 


Similarly,  the  area  may  be  unbounded 
at  the  left  hand,  and  the  lower  limit  or  lM^        y/M^ 

both  limits  of  the  definite  integral  may 
be  infinite. 

In  like  manner  let  f(x)  become  infinite  at  the  upper  limit, 
and  the  curve  y  =f(x)  approach  x  =  b  as  an  asymptote.  Then 
the  area  bounded  by  the  curve,  the  axis  of  x,  an  ordinate  x  =  a, 
and  an  ordinate  near  the  asymptote  x  =  b  may  approach  a 
definite  value  as  the  latter  ordinate  approaches  the  asymptote. 

Such  an  area  may  be  expressed  by  the  integral 

Lim  f      f(x)  dx  =  Lim  F(b  -  A)  -  F(a), 


or,  more  concis 


3ely>  f  /O) 


dx  =  F(b)  -  F(a). 


Ex.  4 


J"a       dx  T  .       ,xla 

=    sin-1- 
o   Va2  —  x2      L  "-1° 


MEAN  VALUE 
..       (Fig.  163) 


265 


a    - 
Fig.  1G4 

Failure   to  do  this 


(Fig.  164) 
Similarly,  f(x)  may  become  infi- 
nite at  the  lower  limit  or  at  both 
limits.  If  it  becomes  infinite  for 
any  value  c  between  the  limits, 
the  integral  should  be  separated 
into  two  integrals  having  c  for  the 
upper  and  the  lower  limit  respectively 
may  lead  to  error. 

X+ldx 
—  ■ 
i     ./■'- 

Since — becomes  infinite  when  x  =  0  (fig.  165),  w 

x'1 
separate  the  integral  into  two,  thus: 

J-i    x-      J -\  x-      Jo    .i- 
Had  we  carelessly  applied  the  incorrect  formula 

xrs-t-as 

we  should  have  been  Led  to  the  absurd  result  -  2.  FlG-  165 

127.  The    mean   value   of    a   function.     In    fig.  166    let   the 
curve  DPC  be  the  graph  of  the  function /(a;).    Then 


r 


f(x)  dx  =  area  ADPCB, 


Let  m  =  AX  and  M=AH 
be  respectively  the  smallest 
and  the  largest  value  assumed 
by  f(x)  in  the  interval  AB. 
Construct  the  rectangle  A  BKH 
with  the  base  AB  and  the  alti- 
tude AM—  M.  Its  area  is  AB  •  AH=(b  -  a)M.  Construct  also 
the  rectangle  ABLN with  the  base  AB  and  the  altitude  AN—  m. 
Its  area  is  AB  •  AN=  (6  —  a)m. 


260  APPLICATIONS  OF  INTEGRATION 

Now  it  is  evident  that  the  area  A  BCD  is  greater  than  the 
area  ABLN  and  less  than  the  area  ABKH.    That  is, 

(b  -  a)  m  <  f  f(x)  dx  <  (b  -  a)  M* 
Consequently  I    f(x)  dx  =  (b  —  a)  /x, 

U  a 

where  fx,  is  some  quantity  greater  than  m  and  less  than  M,  and 
is  represented  on  fig.  166  by  AS.  But  since  f(x)  is  a  continuous 
function,  there  is  at  least  one  value  £  between  a  and  b  such 
that  /(|)  =  fr  anfl  therefore 


f. 


f(x)dx  =  (b-d)f(g).  (1) 


Graphically,  this  says  that  the  area  ABCD  is  equal  to  a  rec- 
tangle ABTS  whose  base   is  AB  and  whose  altitude  AS  lies 
between  AN  and  AH. 
From  (1)  we  have 

/(D=^J^/(*)^  (2) 

where  f  lies  between  a  and  b.    The  value 


l      rh 


</./• 


is  called  the  weaw  i>a?we  of  /(#)  in  the  interval  from  a  to  b.  This 
is,  in  fact,  an  extension  of  the  ordinary  meaning  of  the  average, 
or  mean,  value  of  n  measurements.  For  let  y0,  yv  y.2,  •  •  ■,  yn_x 
correspond  to  n  values  of  x,  which  divide  the  interval  from  a 
to  b  into  n  equal  parts,  each  equal  to  Ax.  Then  the  average  of 
these  n  values  of  y  is 

yo  +  ^+^4-  •  ••  +  ?/„_! 
n 
This  fraction  is  equal  to 

Oo  +  ffi+yaH +yn-,)Ax  =  y^x+y.Ax-hy^Ax-l +  yn_,Ax 

nAx  b  —  a 

*A  slight  modification  is  here  necessary  if  f(x)  =  fc,  a  constant.    Then 
f(x)  dx  =  (b  —  a)  k. 


AREA 


267 


As  n  is  indefinitely  increased,  this  expression  approaches  as  a 

1      rh  1      rb 

limit /    ydx  = I   f(x)  dx.    Hence  the  mean  value 

t>-aJa  h-aJa 

of  a  function  may  be  considered  as  the  average  of  an  "  infinite 
number "  of  values  of  the  function,  taken  at  equal  distances 
between  a  and  b. 

Ex.  1.    Find  the  mean  velocity  of  a  body  falling  from  rest  during  the 

time  tv 

The  velocity  is  gt,  where  g  is  the  acceleration  due  to  gravity.   Hence  the 

1       rh 
mean  velocity  is I     gtdt  —  \gtv   This  is  half  the  final  velocity. 

Ex.  2.    Find  the  mean  velocity  of  a  body  falling  from  rest  through  a 
distance  .s'r 

The  velocity  is  V'2  gs.    Hence  the  mean  velocity  is 

—  fohV2glds  =  %V2^r 

This  is  two  thirds  the  final  velocity. 

128.  Area  of  a  plane  curve  in  polar  coordinates.  Let  0  (fig.  167) 
be  the  pole,  OM the  initial  line  <>f  a  system  of  polar  coordinates 
(r,  0),  OA  and  OU  two  fixed  radius  vectors  for  which  0=  a  and 
B  =  fi  respectively,  and  All  any  curve  for 
which  the  equation  is  r  =/(#).  Required 
the  area  AOH. 

The  required   area  may  be  divided   into 

n    smaller    areas    by    dividing    the     angle 

AOB—^  —  a  into  n  equal  parts, 

6  —  a 

each  of  which  equals  =  A#, 

n 

and  drawing  the  lines  0PX,  OP^ 

0P3,   •  • -,   OPn_v   where  AOP1  = 

%OPi=P2OP3=  •  •  •=Pn_105=A^ 

(In  the  figure  w  =  8.)  The  required 
area  is  the  sum  of  the  areas  of  these 
elementary  areas  for  all  values  of  n.    The  areas  of  these  small 
figures  may  be  found  approximately  by  describing  from  0  as  a 
center  the  circular  arcs  AL\,  2£Ba,  P2P^  •  •  •,  ^_i-K„«    Let 
OA  =  r0,      OP,  =  rv      OP  =  r,,      •  •  • ,      0P_X=  rn  _  v 


268  APPLICATIONS  OF  INTEGEATION 

Then,  by  geometry, 

the  area  of  the  sector  A 0/>\  =  }2  r'^AO, 
the  area  of  the  sector J^OB2=  ^  r'^AO, 

the  area  of  the  sector B_1OBn  =  -* rl_xA6. 
The  sum  of  these  areas,  namely 

is  an  approximation  to  the  required  area,  and  the  limit  of  this 
sum  as  n  is  indefinitely  increased  is  the  required  area.    Hence 


the  area  A  OB 


-*/V* 


The  above  result  is  unchanged  if  the  point  A  coincides  with 
0,  but  in  that  case  OA  must  be  tangent  to  the  curve.  So  also  B 
may  coincide  with  0. 

Ex.    Find  the  area  of  one  loop  of  the  curve  r  =  a  sin  3  6  (fig.  101,  §  60). 

As  the  loop  is  contained  between  the  two  tangents  $  =  0  and  $  =  —  >  the 
required  area  A  is  given  by  the  equation 


1  Jo  2  Jo 


cos  .6  8    a      ira2 
"2 d6=U\ 

129.  Volume  of  a  solid  with  parallel  bases.  Fig.  168  repre- 
sents a  solid  with  parallel  bases.  The  straight  line  Off  is  drawn 
perpendicular  to  the  bases,  cutting  the  lower  base  at  A,  where 
h  =  a,   and  the  upper  base    at  B,   where   h  =  b.    Let  the   line 

AB  be  divided  into  n  parts  each  equal  to  =Ah,  and  let 

n 

planes  be  passed  through  each  point  of  division  parallel  to  the 

bases  of  the  solid.     Let  Ao  be  the  area  of  the  lower  base  of 

the  solid,  A1  the  area  of  the  first  section  parallel  to  the  base, 

A2  the  area  of  the  second  section,  and  so  on,  An_x  being  the 

area  of  the   section  next  below  the   upper  base.    Then  AQAh 

represents  the  volume  of  a  cylinder  with  base  equal  to  Ao  and 

altitude  equal  to  Ah,  A^Ah  represents  the  volume  of  a  cylinder 


VOLT. ME 


269 


standing  on  the  next  section  as  a  base  and  extending  to  the 
section  next  above,  and  so  forth.    It  is  clear  that 

A0Ah  +  A^h  +  J2A7i  +  ...  +  An_xMi  =  j?AtAh 


is  an  approximation  to  the  volume  of  the  solid,  and  that  the 
limit  of  this  sum  as  n  indefinitely  increases  is  the  volume  of 
the    solid.      That    is,    the    required 
volume   /'  is 


V=  C  Adh. 


To  find  the  value  of  this  integral 
it  is  necessary  to  express  .1  in  terms 
of  h,  or  both  A  and  //  in  terms  of  some 
other  independent  variable.  This  is 
a  problem  of  geometry  which  must 
be  solved  for  each  solid.  It  is  clear 
that  the  previous  discussion  is  valid 
if  the  upper  base  reduces  to  a  point, 
i.e.  if  the  solid  simply  touches  a 
plane  parallel  to  its  base.  Similarly, 
both  bases  may  reduce  to  points. 


108 


Ex.  1.    Two  ellipses  with  equal  major  axes  arc  placed  with  their  equal 


•pendicular.    A  variable  ellipse  move 


axes  coinciding  and  their  planes  p 

so    that   the    ends  of    its 

axes  are  on  the  two  given 

ellipses,  the  plane  of  the 

moving  ellipse  being  per- 

pendicular  to  those  of  the 

given   ellipses.     Required 

the   volume   of    the   solid 

generated. 

Let  the  given  ellipses 

be  ABA'W  (fig.  169)  with 

semiaxes  OA  =a  and  OB  =  b,  ami   ACA'C  with   semiaxes  OA  =a  and 

OC=C,  and  let  the  common  axis  be  OX.     Let  NMN'M'  be  one  position 

of  the  moving  ellipse  with  the  center  P  where  OP  —  .r.    Then  ii  A  is  the 

area  of  NMN'M',  n    „     „       ,„ 

A  =  ir  •  PM  ■  PN.  (By  Ex.  1,  §  125) 


169 


270 


APPLICATIONS  OF  INTEGRATION 


But  from  the  ellipse  ABA'B'      -n  +  ^f-  =  1 

a2  lr 

and  from  the  ellipse  ACA'C  -  +  J—^-  =  1 


be 


PM  -PN  =  -(a2-  x2). 
a2 


Therefore 

Consequently  the  required  volume  is 

— —  (a2  —  x2)  dx  =  -  Trdbc. 
■  a  a-  o 

The  solid  is  called  an  ellipsoid  (§  143,  Ex.  5). 

Ex.  2.    The  axes   of    two  equal  right  circular   cylinders   intersect   at 
right  angles.     Required  the  volume  common  to  the  cylinders. 

Let  OA  and  OB  (fig.  170)  be  the  axes  of  the 
cylinders,  OY  their  common  perpendicular  at 
their  point  of  intersection  0,  and  a  the  radius 
of  the  base  of  each  cylinder.  Then  the  figure 
represents  one  eighth  of  the  required  volume  V. 
A  plane  passed  perpendicular  to  OF  at  a  dis- 
tance ON  =  y  from  0  intersects  the  solid  in  a 
square,  of  which  one  side  is 


NP  =  -VOP2-  ON2  =  Va2  -  y2. 
Therefore        \  V=  FnP*  dy  =   f*  (a2  -  y8)  dy  =  §  a8 

Jo  t/0 

and  F=Vl«3- 

130.  Volume  of  a  solid  of  revolution.  A  solid  of  revolution  is  a 
solid  generated  by  the  revolution  of  a  plane  figure  about  an  axis 
in  its  plane.  In  such  a  solid  a  section  made  by  a  plane  perpen- 
dicular to  the  axis  is  a  circle,  or  is  bounded  by  two  or  more 
concentric  circles.  Therefore  the  method  of  the  previous  article 
can  usually  be  applied  to  find  the  volume  of  the  solid.  No 
new  formulas  are  necessary.  The  following  examples  illustrate 
the  method. 

Ex.  1.  Find  the  volume  of  the  solid  generated  by  revolving  about 
OX  the  figure  bounded  by  the  parabola  y2  —  \px,  the  axis  of  x,  and 
the  line  x  =  a. 


VOLUME 


271 


The  area  to  be  revolved  is  shaded  in  fig.  171.  Let  P(x,  y)  be  a  point 
on  the  parabola.  Then  any  section  of  the  solid  through  P  perpendicular 
to  OX  is  a  circle  with  radius  MP  =  y.  Hence  in  the  formula  of  §  129  we 
have  A  =  try2  and  dh  —  dx.  Hence 
the  required  volume  V  is 


■fjirfdx. 


But  from  the  equation  of  the  pa- 
rabola y-  —  4  par.    Therefore 


'•=4WT 


dx 


ipmr 


Ex.  2.    Find  the  volume  generated  by  revolving  around  the  line  x 
the  figure  described  in  Ex.  1. 

If  P  (fig.  172)  is  a  point  on  the  curve, 
a  section  of  the  required  solid  through 
P  and  perpendicular  to  All  is  a  circle 
with  radius  PN  =  u  —  x.  Hence  in  the 
general  formula  of  §  129  A  =  ir  (n  —  x)- 
and  dh  —  dy.  When  x  =  a,  y  =  2-Vpa. 
Hence  the  volume  V  is  given  by 

V  =  f*^~aTr(a  -  xfdy  =  Trf02y/Pl'(«2  -  2  ax  + 

But  from  the  equation  of  the  parabola  x  =  - —    Hence 

Jo         \         2/>      li)ji-/  Id 

Ex.  3.    Find  the  volume  of  the  ring  solid  generated  by  revolving  a  circle 
of  radius  a  about  an  axis  in  its  plane  b  units  from  the  center  (ft  >  a). 

Take  the  axis  of  revolution  as  OY 
(fig.  173)  and  a  line  through  the  center 
as  OX.  Then  the  equation  of  the  circle 
is  (x  -  ft)2  +  y*  =  a'2. 

A  line  parallel  to  OX  meets  the  circle 


!->A>a 


Fm.  172 


16      4   J 


in  two  points,  A  where  x  =  x1  =  ft  —  Va2  —  y2 
and  B  where  x  =  x2=  ft  +  Vo2  —  y2.  A  sec- 
tion of  the  required  solid  taken  through 
AB  perpendicular  to  OY  is  bounded  by 
two  concentric  circles  with  radii  xt 
and    x2    respectively.       Hence    in    §  129 

A  =  TTxf  —  7TJ'2,  and  dh  =  dy.    The  summation  extends  from  the  point  L 
where  y  ——  a  to  the  point  A'  where  y  =  +  a.    Hence,  for  the  volume  V, 


T  =  ^ J_    (''-" _  'ri") dy  =  4  irhi_ ,  x     _  ll~ dy 


-a-h. 


APPLICATIONS  OP  INTEGRATION 


131.  Length  of  a  plane  curve.  To  find  the  length  of  any 
curve  AB  (fig.  174),  assume  n  —  1  points,  //,  7f,  •  •  •,  %_v  be- 
tween A  and  /.'  and  connect  each  pair  of  consecutive  points  by 
a  straight  line.  The  length  of  AB  is 
then  defined  as  the  limit  of  the  sum 
of  the  lengths  of  the  n  chords  APX, 
I^B,,  P,PZ,  •  •  .,  Pl_1B  as  n  is  increased 
without  limit  and  the  length  of  each 
chord  approaches  zero  as  a  limit.  By 
means  of  this  definition  we  have 
already  shown  (§§  91  and  104)  that 


Fig.  174 


d8  =  Vda?+dtf 

in  Cartesian  coordinates,  and 


in  polar  coordinates. 
Hence  we  have 

and 


ds  =  ^dr+r\W> 

8=fV,h*+d!f 

8=  fy/d)*+i*dff\ 


CO 

(2) 
(3) 
(4) 


To  evaluate  either  (3)  or  (4)  we  must  express  one  of  the 
variables  involved  in  terms  of  the  other,  or  both  in  terms  of  a 
third.    The  limits  of  integration  may  then  be  determined. 

It  may  be  noticed  that  (4)  can  be  obtained  from  (3).    For 

we  have  /.  •    a 

x  —  r  cos  6,         y  =  r  sm  0. 

Then  dx  =  cos  6  dr  —  r  sin  6  dQ, 

dy  =  sin  6  dr  +  r  cos  6  dd, 

and  dx2  +  dy1  =  dr*  +  r\W\ 

Ex.  1.  Find  the  length  of  the  parabola  if  =  ^  Px  from  the  vertex  to 
the  point  (h,  I). 

From  the  equation  of  the  parabola  we  find  2  ydy  =  ipdx.  Hence 
formula  (3)  becomes  either 


t/U     \  ±p-  1  p  J  a 


LENGTH  273 


Either  integral  leads  to  the  result 


*  =  —  Vi2  +  4pa  +  p  log !—■ 

ip  2p 

Ex.  2.    Find  the  length  of  the  epicycloid  from  cusp  to  cusp. 
The  equations  of  the  epicycloid  are  (§  57) 

x  =  (a  +  1>)  cos  <f>  —  a  cos <f>, 

y  =  ("  +  '0  sin  <f>  —  a  sin  — (j>. 


Ileuce 


Then 


dx  =  \-  (a  +  6)  sin  <£  +  («  +  &)  sin  ^— ■  <^>1  ,/<£, 
rfy=  ["(„  +  5)  cos  <£-(«  +  &)  cos  ^-±-^lrf<£. 


/,•  =  (a  +  6)  y-J  -  2  /sin  0  sin  i±i  <£  +  c< is  <£  c< >s  5!±*  ^  ,/<£ 
=  («  +  6)  \/2  -  2  cos  -  <M£  =  2  (a  +  6)  sin  —  <£  ,/<£. 
Therefore      s  =  2  (a  +  6)  J^  *   sin  _''   tft  ,/<f>  =  —  (a  +  &). 

132.  The  work  of  the  previous  article  may  be  brought  into 
connection  with  §124  as  follows: 


kT7%y 


Since        y_(^  +  c^^_N       ^A,/A, 


V  (/./"  +  (/// 


>HBr 


|1+ /%y 

,.             T.     V(A.»-)'-+(A//)a      ..     N        VA./7  A./-     1 

then       Lim     v      7 ^=^-  =  Lnn  — -  Lim  —  ==  1. 


+m 


v(/.r  +  J/z- 
and  V(A.r)2  +  (Aty)2  =  Vj.r  +  ,ty2  +  e  V^  +  df- 


By  §124  the  term  eV,lr  +  thf  will  not  affect  the  limit  of 
2V(A,-)2  +  (A^. 


^ 


7Ps 


7p  4 

1  8 


Fig.  175 


274  APPLICATIONS  OF  INTEGRATION 

133.  Area  of  a  surface  of  revolution.  A  surface  of  revolution 
is  a  surface  generated  by  the  revolution  of  a  plane  curve  around 
an  axis  in  its  plane  (§  130).  Let  the  curve  AB  (fig.  175) 
revolve  about  OH  as  an  axis.  To  find  the  area  of  the  surface 
generated,  assume  n  —  1  points,  //,  Pv  P6,  •  •  ., 
i£_u  between  A  and  B  and  connect  each 
pair  of  consecutive  points  by  a  straight  x6 
line.  These  lines  are  omitted  in  the  figure  -^ 
since  they  are  so  nearly  coincident  with  the  iV4 
arcs.  The  surface  generated  by  AB  is  then  n2 
defined  as  the  limit  of  the  sum  of  the  areas      ^' 

No 

of  the  surfaces  generated  by  the  n  chords 
APV  B^P,,  i£j£,  •  •  .,  Pn_xB  as  n  increases 
without  limit  and  the  length  of  each  chord 
approaches  zero  as  a  limit. 

Each  chord  generates  the  lateral  surface  of  a  frustum  of  a 
right  circular  cone,  the  area  of  which  may  be  found  by 
elementary  geometry. 

Draw  the  lines  AJSf0,  I{NV  B^N^  •  •  •  perpendicular  to  OH,  and 

Place      N0 A  =  rD,  NXPX  =  rv  K2P2  =  rif .  •  -,  A^J^  =  ra. 

Then  the  frustum  of  the  cone  generated  by  PPt  +  x  has  for  the 
radius  of  the  upper  base  2fi+1J^+v  for  the  radius  of  the  lower  base 
NtP,  and  for  its  slant  height  I*Pi+v  Its  lateral  area  is  therefore 
equal  to  (N.P+N.     P    ^ 

Therefore  the  lateral  area  of  the  frustum  of  the  cone  equals 

This  is  an  infinitesimal  which  differs  from 

2  wr$s 

by   an   infinitesimal   of   higher   order,    and   therefore   the   area 

generated  by  AB  is  the  limit  of  the  sum  of  an  infinite  number 

of  these  terms.    Hence,  if  we  represent  the  required  area  by  S, 

we  have  r 

S=2tt     rds. 


WORK  275 

To  evaluate  the  integral  it  is  necessary  to  express  r  and  ds  in 
terms  of  the  same  variable  and  supply  the  limits  of  integration. 

Ex.    Find  the  area  of  the  surface  of  revolution  described  in  Ex.  1,  §  130. 

Here   r  =  y  and  ds  =  Vdx2  +  dy2,  where  x  and  y  satisfy  the  equation 

y"  =  ipx.    Consequently  we  may  place  r  =  2  Vpx,  and,  as  in  Ex.  1,  §  131, 

ih  _     /*  +  P  dj. 
Then  S  =  4  tt  VJ,  f"  Vx  +  p  dx  =  §wVp[(a  +  p)*  -/<-]• 

134.  Work.  By  definition,  the  work  done  in  moving  a  body 
against  a  constant  force  is  equal  to  the  force  multiplied  by  the 
distance  through  which  the  body  is  moved.  Suppose  qow  that 
a  body  is  moved  along  OX  (fig.  17G)  from  A  (x  —  a)  to  B  (x  =  /<) 
against  a  force  which  is  not  -  ,  .  ,  t  ,  .  «  ,  y 
constant   but    a   function    of   a;  A  5S  *  M*  *  *  U 

and  expressed  by/(:r).    Let  the  Fl°"  176 

line  AB  be  divided  into  n  equal  intervals,  each  equal  to  Ax,  by 
the  points  M^  M%,  Mz,  .  .  •,  Mn_x.    (In  fig.  176,  n  =  7.) 

Then  the  work  done  in  moving  the  body  f rom  A  to  Mi  would 
be  f  (a*)  Ax  if  the  force  were  constantly  equal  to  /(«)  through- 
out the  interval  AM.  Consequently,  if  the  interval  is  small, 
/(rt)Ax  is  approximately  equal  to  the  work  done  between  A 
and  Mx.  Similarly,  the  work  done  between  Ml  and  Ma  is  approxi- 
mately equal  to/(^)  A.r,  that  between  JA,  and  3fg  approximately 
equal  to/(:r,)A:r,  and  so  on.  Hence  the  work  done  between  A 
and  B  is  approximately  equal  to 

f(a)Lx+f(xdLx+f(x^*x+.  ..+f(zn_JAz. 

The  larger  the  value  of   n,  the  better  is  this  approximation. 
Hence  we  have,  if  W represents  the  work  done  between  A  and  /.', 


\Y  - 


Lim£ /(.•/)  A.r  =  Cf(x)dx. 


135.  Pressure.  Consider  a  plane  surface  of  area  A  immersed 
in  a  liquid  at  a  uniform  depth  of  h  units  below  the  surface. 
The  submerged  surface  supports  a  column  of  liquid  of  volume 
hA,  the  weight  of  which  is  whA,  where  w  (a  constant  for  a 
given  liquid)   is  the   weight  of  a  unit  volume   of  the  liquid. 


270  APPLICATIONS  OP  INTEGRATION 

This  weight  is  the  total  pressure  on  the  immersed  surface.  The 
pressure  per  unit  of  area  is  then  wh,  which  is  defined  as  the 
pressure  at  a  point  h  units  below  the  surface.  By  the  laws  of 
hydrostatics  this  pressure  is  exerted  equally  in  all  directions. 
We  may  accordingly  determine,  in  the  following  manner,  the 
pressure  on  plane  surfaces  which  are  perpendicular  to  the 
surface   of   the   liquid: 

Let  BRQ  (fig.  177)  be  a  plane  surface  so  immersed  that  its 
plane  is  perpendicular  to  the  surface  of  the  liquid  and  inter- 
sects that  surface  in  the  line  TS.    Divide  BRQ  into  strips  by 
drawing  lines  parallel  to  TS.    Let 
the    depth   of    a    line   of    the   first 
strip  be  hQ,  that  of  the  second  strip  ^^M^^ 

be  h^  that  of  the  third  strip  be  7;.,,  /  =\ 

and  so  on.     Call  the   area   of   the 


first  strip  (A^)0,  that  of  the  second  /  \p 

strip  (A.i)^  that  of  the  third  strip  FlG  177 

(A.4)2,  and  so  on.    Then  the  pres- 
sure on  the  first  strip  is  approximately  wh0(AA)0,  that  on  the 
second  strip  is  approximately  u'h^AA)^  that  on  the  third  strip 
is  approximately  w\(AA)^  etc.     Therefore  the  total  pressure 
on  BRQ  is  approximately 
w  [K(AA)0  +  hiAA)^  .  • .  +  hB_1(^AAX-t']  =  u>X  ^(A^),.. 

This  approximation  is  better  the  greater  the  number  of  strips, 
since  we  have  taken  the  whole  strip  as  lying  at  the  level  of 
the  same  line.  Therefore  the  total  pressure  P  is  the  limit  of  the 
above  sum  as  n  =  oo  ;  that  is, 


"="'/ 


lid  A. 


To  evaluate  the  integral  it  is  necessary  to  express  h  and  A  in 
terms  of  the  same  variable  and  supply  the  limits.  In  finding  dA 
the  strips  may  be  taken  as  rectangles,  as  in  finding  the  area. 

Ex.  A  parabolic  segment  with  base  2  h  and  altitude  a  is  submerged  so 
that  its  base  is  in  the  surface  of  the  liquid  and  its  axis  vertical. 

Let  I?QC  (fig.  178)  be  the  parabolic  segment  and  let  CB  be  drawn 
through  the  vertex  of  the   segment  perpendicular  to    TS.    According  to 


CENTER  OF  PRESSURE 


277 


the  data  RQ  =  2b,  CB  =  a.  Draw  a  horizontal  strip  LXA\LV  with  its 
bottom  line  cutting  CB  at  .1/.  Let  CM  =  x ;  then  the  depth  h  of  the  line 
LN  is  a  —  x  and  the  breadth  MMX  of  the  strip  is  dx. 

R  B  Q  * 


dx. 


Consequently  dA 

=  (LN)dx. 

But,  from  §  45, 

LN2      CM. 

RQ2       CB  ' 

whence 

a 

and  therefore 

a2 

Therefore,  since 

x  =  0  at  ( ',  ; 

given  by 

P 

=  u         —  (a 

X) 


Fig.  178 
=  a  at  B,  the  total  pressure  P 

.,-ilx  =  —  irhir. 

15 


136.  Center  of  pressure.  From  mechanics  we  take  the  fol- 
lowing principles: 

1.  The  resultant  of  a  set  of  parallel  forces  is  equal  to  the 
sum  of  the  forces. 

2.  The  moment  of  a  force  about  a  line  al  righl  angles  to  the 
line  of  action  of  the  force  is  defined  as  the  product  of  the  force 
and  the  shortest  distance  between  the  two  lines. 

3.  The  moment  about  a  line  of  the  resultant  of  a  number  of 
forces  is  equal  to  the  sum  of  the  moments  of  the  forces. 

Now  in  the  pressure  problem  of  §  135,  the  pressure  on  each 
one  of  the  elementary  strips  is  a  force  approximately  equal  to 
whAA  acting  at  right  angles  to  the  area.  By  the  second  principle 
stated  above,  the  moment  of  this  force  about  TS  is  h(tchAA), 
and  the  limit  of  the  sum  of  the  moments  of  all  the  forces  is 


fh(whdA)  =  w  fir  J  A 


By  the  first  principle  stated  above,  the  resultant  of  the  pres- 
sures on  all  the  rectangles  is  the  total  pressure  P.  If  this  acts 
at  a  distance  h  below  the  surface  of  the  liquid,  we  have,  by  the 
third  principle,  _ 

hP  =  w      JcJA, 

from  which  h  can  be  found. 

AC 


278  APPLICATIONS  OF  INTEGRATION 

The  point  at  which  P  acts  is  called  the  center  of  pressure. 

The  formula  above  gives  the  depth  of  the  center  of  pressure- 
Ex.    Find  the  depth  of  the  center  of  pressure  of  the  parabolic  segment 

of  the  example  in  §  135. 

From  the  discussion  just  given, 

2  b  .  v„  i  ,        32mj6o8 


—  (a  —  x)2x^c?a 


Ph 


F 


But  P  =  T85  wba?  (Ex.,  §  135).  Therefore  h  =  \ a.  By  symmetry  the 
center  of  pressure  lies  in  CB,  and  is  therefore  fully  fixed. 

137.  Center  of  gravity.  Consider  n  particles  of  masses  mx, 
m2,    ms,  ••-,  mn,    placed    at    the    points   P[(xv  ^),  i£02,  #2), 

50«v  y.).  •-.  ^fe  y.)  (%• 179)  re- 

spectively.  The  weights  of  these  particles 
form  a  system  of  parallel  forces  equal  to 
m-yff,  m2g,  m3g,  •  .  .,  mng,  where  g  is  the 
acceleration  due  to  gravity.  The  principles 
of  mechanics  stated  in  §  136  are  therefore 
applicable.  The  resultant  of  these  forces  is 
the  total  weight  W  of  the  n  particles,  where 

i—n 

W  =  mxg  -f  m„g  +  mzg  -f- .  . .  +  mng  =  9^  m>i- 

This  resultant  acts  in  a  line  which  is  determined  by  the  con- 
dition that  the  moment  of  W  about  any  line  through  0  is  equal 
to  the  sum  of  the  moments  of  the  n  weights. 

Suppose  first  the  figure  placed  so  that  gravity  acts  parallel 
to  OY,  and  that  the  line  of  action  of  W  cuts  OX  in  a  point 
the  abscissa  of  which  is  x.  Then  the  moment  of  W  about  a 
line  through  0  perpendicular  to  the  plane  XOY  is  gx^mp  and 
the  moment  of  one  of  the  n  weights  is  gmfr. 

Hence  gx  V  mi  =  g  V  m^. 

Similarly,  if  gravity  acts  parallel  to  OX,  the  line  of  action  of 
the  resultant  cuts  OF  in  a  point  the  ordinate  of  which  is  y, 
where  _^  ^ 


CENTER  OF  GRAVITY  279 

These  two  lines  of  action  intersect  in  the  point  G,  the  coordi- 
nates of  which  are       __<  ^ 

*  =  %— '         3/  =  %—  C1) 

Z,  mi  Zf  ^ 

Furthermore,  if  gravity  acts  hi  the  XO  Y  plane,  but  not  paral- 
lel to  either  OX  or  0  Y,  the  line  of  action  of  its  resultant  always 
passes  through  G.  This  may  be  shown  by  resolving  the  weight  of 
each  particle  into  two  components  parallel  to  OX  and  OY  respec- 
tively, finding  the  resultant  of  each  set  of  components  in  the 
manner  just  shown,  and  then  combining  these  two  resultants. 

If  gravity  acts  in  a  direction  not  in  the  XO  V  plane,  it  may  still 
be  shown  that  its  resultant  acts  through  G,  but  the  proof  requires 
a  knowledge  of  space  geometry  not  yet  given  in  this  course. 

The  point  G  is  called  the  center  of  gravity  of  the  »  'particles. 

If  it  is  desired  to  find  the  center  of  gravity  of  a  physical  body, 
the  solution  of  the  problem  is  as  follows:  The  body  in  question 
is  divided  into  n  elementary  portions  such  that  the  weight  of 
each  may  be  considered  as  concentrated  at  a  point  within  it. 
If  m  is  the  total  mass  of  the  body,  the  mass  of  each  element 
may  be  represented  by  Am.  Then  if  (>.  //, )  arc  the  coordinates 
of  the  point  at  which  the  mass  of  the  ith  element  is  concentrated, 
the  center  of  gravity  of  the  body  is  given  by  the  equations 

V.r.Am  V  if  Am 

*  =  Lim^- — ,  y=Um^r    — ; 

Z/Am  2j  Am 

I  xdm  |  ydm 

whence  *  =  ~~r — "'  ^  =  ~7 '  ® 

I  dm  I  dm 

To  evaluate,  the  integrals  must  be  expressed  in  terms  of  a 
single  variable  and  the  limits  supplied. 

It  is  to  be  noticed  that  it  is  not  necessary,  nor  indeed  always 
possible,  to  determine  x9  y{  exactly,  since,  by  §  124, 

Lim  V  (xt  -f  e.)  Am  =  Lim  Va;.Am, 
»=«  «  =  i  »<  =  *  ,-=i 

if  e,  approaches  zero  as  Am  approaches  zero. 


280 


APPLICATIONS  OF  INTEGRATION 


Ex.  1.    Find  the  center  of  gravity  of  a  quarter  circumference  of  the 
circle  at2  +  ?/2  =  a2,  which  lies  in  the  first  quadrant. 

Let   the   quarter    circumference    be    divided    into   elements   of    arc   ds 
(fig.  180)  ;  then,  if  p  is  the  amount  of  mass  per  unit  length, 
dm  =  p  ds. 

The  mass  of  each  element  may  be  considered  concentrated  at  a  point 
(x,  y)  of  the  curve.    Hence 

Jpxds    .  fpjjds 

y 


fpds  ff 


ds 


If  p  is  assumed  constant,  it  may  be  removed   from   under  the   integral 
signs   and   canceled.     The   denominator   of    each    fraction  is   then   equal 
to  s,  a  quarter  circumference.     To  compute 
the  numerators,  we  have,  from  the  equation 
of  the  curve, 


ds  =  Vdx2  +  df  =  -  dx 

y 


dy, 


where  s  is  assumed  as  measured  from  A  so 
that  dx  is  positive  and  dy  negative. 


Therefore 


and 


Hence 


f  xds  =  —  f   a  dy  —  a2, 
|  ij  ds  =    f  a  dx  =  «2, 

/  ds  =  — ,  a  quarter  circumference. 

_      _      2  a 
x  =  y  = 


Ex.  2.  Find  the  center  of  gravity  of  a  quarter  circumference  of  a  circle 
when  the  amount  of  matter  in  a  unit  of  length  is  proportional  to  the  length 
of  the  arc  measured  from  one  extremity. 

As  in  Ex.  1,  dm  =  pds,  but  here  p  =  ks,  k  being  a  constant.    Then 
dm  =  ksds. 

The  integration  is  best  performed  by  use  of  the  parametric  equations  of 
the  circle  (§  53).    Then 

/sxds        (  2  a3(j>  cos  <b  dd>         , 
Jo  (*  t  —  o)a 


f»*       rh*<j,d<p 

Jo 
j'sy  ds        C  2  a*cp  sin  cf>  d<p       g  fl 


CENTER  OF  GRAVITY 


281 


Ex.  3.  Find  the  center  of  gravity  of  the  area  bounded  by  the  parabola 
y"  =  1  px  (fig.  181),  the  axis  of  x,  and  the  ordinate  through  a  point  (/*,  h) 
of  the  curve. 

As  in  finding  the  area,  let  the  area  be  divided  into  elementary  rectangles 
ydx,  where  (x,  y)  is  a  point  on  the  curve.  Then,  if  p  is  the  amount  of  mass 
per  unit  area,  dm  =  pydx, 

and  this  mass  may  be  considered  as  concentrated  at  the   middle  point 


Then 


of  its  left-hand  ordinate. 


X 


py 


dx 


jQpydx 


If  p  is  assumed  constant,  it  may  be  removed  from   under  the  integral 
signs,  and  canceled.  Then,  by  aid  of  the  equation  of  the  curve,  we  compute 

the  integrals 


and 


fhxydx  =2p*  (hA,lx  =  *p*A*=  I  Irk, 

Jo  Jo 

i(  f  y*dx  =  2p  f  xdx  =  ph*  =  1  /</.-, 
"Jo  Jo 

f  y  dx  =  2j,}-  f  x*dx  =  §  pM  =  §  hk. 
x=:&h,         y  =  |  k. 


Therefore  x  =  ;'-;  A, 

Ex.  4.  Find  the  center  of  gravity  of  the  segment  of  the  ellipse  —  + 


(fig.  182)  cut  off  by  the  chord  through   the  positive  ends  of  the  axes  of 
the    curve.      Divide    the    area    into    elements    by  j 

lines  parallel  to  OY.  If  we  let  ya  lie  the  ordi- 
nate of  a  point  on  the  ellipse,  and  y1  the 
ordinate  of  a  point  on  the  chord,  we  have 
as    the    element    of    area, 


and  hence 


(ya-yi)<fe, 

dm  =  p(y,  -yjdx, 


Fig.  182 

where  p,  the  amount  of  mass  per  unit  of  area,   is  assumed  constant. 

The  mass  of  this  element  may  be  considered  as  concentrated  at  the 
point  (x,h±h). 

Hence 


//,)./•</./• 


X'<* 


282 


APPLICATIONS  OP  INTEGRATION 


From  the  equation  of  the  ellipse,  y„  =  -  Va2  —  x2 ;  from  the  equation  of 
I  a 

the  chord,  y1  =  —  (a  —  x). 


The  denominator 


J(?/2  —  y{)dx  is  equal  to  the  area  of  the  quadrant 
o       '  *  i  _i 


of  the  ellipse  minus  that  of  a  right  triangle,  i.e.  is  equal  to 


Hence 


0      r«  r— 

-   \    x\Va? 
a  Jo      L 


(a 


■)-]dx 


^p  f0*l(*-'6-(a-*')*]d* 


obi 


8  (»— 2) 


2  6 


8  (—2) 


Ex.  5.  Find  the  center  of  gravity  of  a  spherical  segment  of  one  base 
generated  by  revolving  the  area  BDE  (fig.  183)  about  OY,  where  OB  =  a, 
and  OE  =  c. 

Let  the  volume  be  divided  into  elementary 
cylinders  as  in  §  130.  Then  the  element  of 
volume  is  Ady  =  irx2dy,  and  hence 

dm  =  pTrx2dy, 

where  p  is  the  density,  assumed  constant.  The 
mass  of  this  element  may  be  considered  as  con- 
centrated at  (0,  ?/),  the  center  of  its  base.  Hence 
the  center  of  gravity  of  the  entire  volume  is  in 
the  line  0  Y,  and  its  ordinate  y  is  given  by 

J]  y(jp-ir&dy)      £  («*>J-yz)dy 


Fig.  183 


J    pirx2dy  j    ('r-if)dy 


3  (a  +  c)2 

4  '  2a  +  c  ' 


Ex.  6.    Find  the  center  of  gravity  of  the  surface  of  the  sjmerical  seg- 
ment of  Ex.  5. 

Divide  the  surface  into  elementary  bands  as  in  §  133.    Then 
dm  =  2  irpx  ds, 
where  p,  the  amount  of  mass  per  unit  area,  is  assumed  constant. 

This  mass  may  be  considered  concentrated  at  (0,  y).    Hence,  using  the 

notation  and  the  figure  of  Ex.  5,  we  have  ds  =  — - ,  and  therefore 

x 

_       r^dS       fj'd>->       a  +  c 


Cxds    J> 


ATTRACTION 


283 


138.  Attraction.    Two  particles  of  matter  of  masses  m1  and 
m2  respectively,  separated  by  a  distance  r,  attract  each  other 

with  a  force  equal  to  k    l ,  2>  where  k  is  a  constant  which  de- 
r 

pends  upon  the  units  of  force,  distance,  and  mass.    We  shall 

assume  that  the  units  are  so  chosen  that  £=1. 

Consider  now  n  particles  of  masses  m^  m2,  mg,  •  •  •,  mn  lying 


in  a  plane  at  the  points  Pv  P, 
be  required  to  find 
their  attraction  upon 
a  particle  of  unit  mass 
situated  at  a  point  A 
in  their  plane. 

Let  the  distances  APV 
AP,  •  •  • ,  AP^  be  denoted 
by  rx,  r2,  •  •  •,  rn.  The 
attractions  of  the  indi- 
vidual particles  are 


PH  (rig.  184).     Let  it 


Fig.  184 


but  these  attractions  cannot  be  added  directly,  since  they  are 
not  parallel  forces.  To  find  their  resultant  we  will  resolve 
each  into  components  along  two  perpendicular  axes  AX  and  AS 
respectively.  If  we  denote  the  angle  XAPt  by  #,,  we  have  as 
the  sum  of  the  components  along  AX, 


X=^cos01+^cos02  + 

r'r  x      r': 


+  -\  cos  en 

r: 


and  for  the  sum  of  the  components  along  AY, 


Wl- 


sin  0  H — -  sin  0„  + 


4-  _?  sin  0n. 


The  resultant  attraction  is  then 


R=Vx*+  Y2 
and  acts  in  a  direction  which  makes  tan" 


J 


with  AX. 


281 


APPLICATIONS  OF  INTEGRATION 


Let  it  now  be  required  to  find,  the  attraction  of  a  material 
body  of  mass  m  upon  a  particle  of  unit  mass  situated  at  a 
point  A.  Let  tlie  body  be  divided  into  n  elements,  the  mass  of 
each  of  which  may  be  represented  by  Am,  and  let  Jf  be  a  point 
at  which  the  mass  of  one  element  may  be  considered  as  concen- 
trated.   Then  the  attraction  of  this  element  on  the  particle  at  A 

is  — -  >  where  rt  =  I?  A,  and  its  component  in  the  direction  AX 

is  — -  cos  6{,  where  6i  is  the  angle  XAI?.  The  whole  body,  there- 
fore, exerts  upon  the  particle  at  A  an  attraction  whose  com- 
ponent in  the  direction  AX  is 

Ajos0 


Lim2^HA™ 


dm. 


Similarly,  the  component  in  the  direction  AY  is 

Ex.    Find  the  attraction  of  a  uniform  wire  of  length  I  and  mass  m 
on  a  particle  of  unit  mass  situated  in  a  straight  line  perpendicular  to 
the  wire  at  one  end,  and  at  a  distance 
a  from  it. 

Let    the    wire    OL    (fig.  185)    be 
placed   in   the    axis    of   y   with    one 

end   at   the   origin,  and  let  the  par-  / 

tide  of  unit  mass  be   at  A  on  the  // 

axis  of  x  where  A  O  =  a.    Divide  OL  //','' 

into  n  parts,  0MV  MXMV  M2M8,  ■■■,  ///'/'' 


Mn  _  i  L,  each  equal  to  -  =  Ay.    Then, 

if  p  is  the  mass  per  unit  of  length  of 
the  wire,  the  mass  of  each  element 
is  Am  =  pAy.  We  shall  consider  the 
mass    of 


Fig.  185 


-X 


each    element    as    concen- 
trated at  its  first  point,  and  shall  in  this  way  obtain  an  approximate 
expression   for   the   attraction    due    to   the    element,  this   approximation 
being  the  better,  the  smaller  A//  is  made.    The  attraction  of  the  element 
M{Mi  +  1  on  A  is  then  approximately 

pAy  _     pAy 
AM?      «2  +  Vl 


where  y.  =  0M(. 


PROBLEMS  285 

The  component  of  this  attraction  in  the  direction  OX  is 

pAy             r\A-h/r             p"A// 
^     '     cos  O A  M.  =  — " - — - , 

H"  +  ^  («* +  *,')* 

and  the  component  in  the  direction  OY  is 

f**    sin  QAM.  =     PyAy     ■ 

Then,  if  A'  is  the  total  component  of  the  attraction  parallel  to  OX,  and 
Y  the  total  component  parallel  to  OY,  we  have 

X  =  Limy1     Pr,A//      =  pa  fl       <h-<        , 
— «ft  (-  +  //f)f  J*  (*  +  ?)* 

F=Lim=^1^^--^- 


y  _jpyAy__    r     y»// 

h   (<r +  !,:)*      PJ°  (>r  +  f)1 


To  evaluate  the  integrals  for  A'  and    )',  place  y  =  atan#.    Then,  if 
a  =  tan-1-  =  OA  /-, 


A'  =  -   I     co% Odd  =  -  sina=     .sin  a, 
a  Jo  a  id 

Y=~   f   Bin  0i/6--(l-  cos  a)  =  -'  (1  -  cos  a), 


since  /p  =  ?n. 

If  R  is  the  magnitude  of  the  resultant  attraction  and  /i  tin-  angle  which 
its  line  of  action  makes  with  OX, 

B  =  VP+P= sin  -  <r, 

«/  2 

B  =  tan-1  —  =  tan-1— =  -  a. 

X  sin  a  2 


PROBLEMS 

1.  Find  the  area  of  an  arch  of  the  curve  y  =  sin  x. 

2.  Find  the  area  bounded  by  the  portions  of  the  curves  y  =  ^sin2cc 
and  y  =  sin  x  +  h  sin  2  x  that  extend  between  a-  =  0  and  x  =  ir. 

3.  Find  the  area  of  the  three-sided  figure  bounded  by  the  coordi- 
nate axes  and  the  curve  x"1  ■+-  y*  =  a*. 

4.  Find  the  area  bounded  by  the  catenary  i/  =  -(e"-{-e  a),  the 
axis  of  x,  and  the  lines  x  =  ±h.  g 

5.  Find  the  area  included  between  the  witch  ?/  =  -5 : — 5  and  its 

asymptote. 


286  APPLICATIONS  OF  INTEGRATION 

6.  Find  the  area  of  one  of  the  closed  figures  bounded  by  the 
curves  y2  =  16  x  and  y2  =  xs. 

7.  Find  the  area  bounded  by  the  curve  y  {x2  +  4)  =  4  (2  —  x),  the 
axis  of  x,  and  the  axis  of  y. 

8.  Find  the  area  bounded  by  the  curve  y2  =  x  (log  a-)2,  the  axis 
of  x,  and  the  ordinates  x  =  1  and  x  =  e. 

9.  Find  the  area  bounded  by  the  parabola  y2  =  2  (x  —  4)  and  the 
line  x  =  3  y. 

10.  Find  the  area  between  the  parabola  a-2  =  4  «y  and  the  witch 
8  a3 

y~*2  +  4a2' 

11.  Find  the  area  bounded  by  the  parabola  x2  —  9y  =  0  and  the 
line  x  —  3//  + 6  =  0. 

12.  Find  the  area  included  between  the  parabolas  y2  =  ax  and 
x2  =  by. 

13.  Find  the  area  bounded  by  the  curve  x2y2  -\-  a2b2  =  a2y2  and  its 
asymptotes. 

14.  Find  the  area  bounded  by  the  hyperbola  —%  —  jz  =  1  and  the 
chord  x  =  h. 

15.  Find  the  area  bounded  by  the  curve  y2(x2  -j-  a2)  =  a2x2  and  its 
asymptotes. 

16.  Find  the  total  area  of  the  curve  81  y2  -f-  4  xi  =  36  x2. 

17 .  Find  the  area  of  the  loop  of  the  curve  (y  —  l)2  =  (x  — 1)2(4  —  x). 

18.  Find  the  area  of  the  loop  of  the  curve  cy2=  (x  —  a)(x  —  b)2, 
(a  <  b). 

19.  Find  the  area  of  the  loop  of  the  curve  16  a3y2=  b2x2(a  —  2x). 

X   ((t  -\-  X^i 

20.  Find  the  area  of  the  loop  of  the  strophoid  y2=  - — -• 

21.  Find  the  area  of  a  loop  of  the  curve  y2(a2  +  x2)  =  x2(a2  —  x2). 

22.  Find  the  total  area  of  the  curve  a2y2  =  xs(2  a  —  x). 

23.  Find  the  area  of  the  loop  of  the  curve  (2  x  +  y)"  =  x2(2  —  x). 

24.  Find  the  area  between  the  axis  of  x  and  one  arch  of  the  cycloid 
X  =  a  (<£  —  sin  <£),  y  =  a  (1  —  cos  <f>). 

25.  Find    the    area    inclosed    by   the    four-cusped    lrypocycloid 
x  =  a  cos30,  y  =  a  sin30. 


PROBLEMS  287 

26.  Find  the  entire  area  bounded  by  the  curve  x  =  acosO, 
y  —  b  sin30. 

27.  Find  the  mean  value  of  the  lengths  of  the  perpendiculars 
from  a  diameter  of  a  semicircle  to  the  circumference,  assuming 
the  perpendiculars  to  be  drawn  at  equal  distances  on  the  diameter. 

28.  Find  the  mean  length  of  the  perpendiculars  drawn  from  the 
circumference  of  a  semicircle  of  radius  a  to  its  diameter,  assuming 
that  the  points  taken  are  equidistant  on  the  circumference. 

29.  Find  the  mean  value  of  the  ordinates  of  the  curve  y  =  sin  x 
between  x  =  0  and  x  =  ir,  assuming  that  the  points  taken  are 
equidistant  on  the  axis  of  x. 

30.  A  number  n  is  divided  into  two  parts  in  all  possible  ways. 
Find  the  mean  value  of  their  product. 

31.  If  the  initial  velocity  of  a  projectile  is  v0,  and  the  angle  of 
elevation  varies  from  0  to  —  >  find  the  mean  value  of  the  range, 
using  the  result  of  problem  36,  Chap.  VII. 

32.  In  a  sphere  of  radius  r  a  series  of  right  circular  cones  is 
inscribed,  the  bases  of  which  are  perpendicular  to  a  given  diameter 
at  equidistant  points.    Find  the  mean  volume  of  these  cones. 

33.  A  particle  describes  a  simple  harmonic  motion  defined  by  the 

equation  8  =  a  sin  kt.    Show  that  the  mean  kinetic  energy  I   -    1 

during  a  complete  vibration  is  half  the  maximum  kinetic  energy  if 
the  average  is  taken  with  respect  to  the  time. 

34.  In  the  motion  defined  in  problem  33,  what  will  be  the  ratio 
of  the  mean  kinetic  energy  during  a  complete  vibration  to  the 
maximum  kinetic  energy,  if  the  average  is  taken  with  respect  to 
the  space  traversed  ? 

35.  Find  the  area  described  in  the  first  revolution  by  the  radius 
vector  of  the  spiral  of  Archimedes  r  =  a  6. 

36.  Show  that  the  area  bounded  by  the  hyperbolic  spiral  rO  =  a 
and  two  radius  vectors  is  proportional  to  the  difference  of  the 
lengths  of  the  radius  vectors. 

37 .  Find  the  total  area  of  the  lemniscate  r2  =  2  a2  cos  2  6. 

38.  Find  the  area  of  a  loop  of  the  curve  r  =  a  sinnfl. 

39.  Find  the  area  of  a  loop  of  the  curve  r2  =  a2  sin  nd. 


288  APPLICATIONS  OF  INTEGRATION 

40.  Find  the  area  swept  over  by  the  radius  vector  of  the  curve 

7j- 

r  =  a  tan  6  as  0  changes  from  0  to  —  • 

41.  Find  the  total  area  of  the  cardioid  r  =  a(l  +  cos  6). 

42.  Find  the  area  of  the  limacon  r  =  2  cos  0  +  3. 

43.  Find  the  area  of  the  curved  strip  of  the  plane  which  has  two 
portions  of  the  initial  line  for  two  boundaries  and  the  arc  of  the 
spiral  r  =  ad  between  6  =  2  ir  and  6  =  6  ir  for  the  other  boundary. 

44.  Find  the  area  of  the  loop  of  the  curve  r2  =  a2  cos  2  6  cos  6 
which  is  bisected  by  the  initial  line. 

45.  Find  the  area  of  a  loop  of  the  curve  r2  sin  6  =  a2  cos  2  6. 

46.  Find  the  area  of  the  kite-shaped  figure  bounded  by  an  arc  of 
a  parabola  and  two  straight  lines  from  the  focus  making  the  angles 
±  «  with  the  axis  of  the  parabola. 

47.  Find  the  area  bounded  by  the  curves  r  =  a  cos  3  6  and  r  =  a. 

4 

48.  Find    the   area   inclosed    by   the   curves   r  = and 

4  J  1  -  cos  e 

1  ~  i  +  cos  e ' 

49.  Find  the  area  cut  off  one  loop  of  the  lemniscate  r2  =  2  a2 cos  2  6 
by  the  circle  r  =  a. 

50.  Find  the  area  of  the  segment  of  the  cardioid  r  =  a(l  +  cos  9) 
cut  off  by  a  straight  line  perpendicular  to  the  initial  line  at  a  distance 
|  a  from  the  vertex. 

51.  Find  the  area  of  the  loop  of  the  curve  (x2  +  y2)3  =  ka2x2y2. 
(Transform  to  polar  coordinates.) 

52.  Find  the  total  area  of  the  curve  (a;2  +  iff  =  4  a?x*  +  4  «y. 

(Transform  to  polar  coordinates.) 

53.  Find  the  area  of  the  loop  of  the  Folium  of  Descartes, 
xz  +  if  —  3  axy  =  0,  by  the  use  of  polar  coordinates. 

x2      if 

54.  On  the  double  ordinate  of  the  ellipse  — :  +  7^  =  1  as  base  an 

isosceles  triangle  is  constructed  with  its  altitude  equal  to  the  dis- 
tance of  the  ordinate  from  the  center  of  the  ellipse  and  its  plane 
perpendicular  to  the  plane  of  the  ellipse.  Find  the  volume  gener- 
ated as  the  triangle  moves  along  the  axis  of  the  ellipse  from 
vertex  to  vertex. 


PROBLEMS  289 

55.  Find  the  volume  cut  from  a  right  circular  cylinder  of  radius 
a  by  a  plane  through  the  center  of  the  base  making  an  angle  6  with 
the  plane  of  the  base. 

56.  Two  parabolas  have  a  common  vertex  and  a  common  axis  but 
lie  in  perpendicular  planes.  An  ellipse  moves  with  its  center  on  the 
common  axis,  its  plane  perpendicular  to  the  axis,  and  its  vertices  on 
the  parabolas.  Find  the  volume  generated  when  the  ellipse  has 
moved  to  a  distance  h  from  the  common  vertex  of  the  parabolas. 

v  57.  An  equilateral  triangle  moves  so  that  one  side  has  one  end  in 
OY  and  the  other  end  in  the  circle  x2  +  f  =  a2,  the  plane  of  the 
rectangle  being  perpendicular  to  OY.  Required  the  volume  of  the 
solid  generated. 

58.  In  a  sphere  of  radius  a  find  the  volume  of  a  segment  of  one 
base  and  altitude  h. 

59.  Find  the  volume  of  the  solid  generated  by  revolving  about  OY 
the  plane  surface  bounded  by  OY  and  the  hypocycloid  xl  +  y*  =  «». 

60.  Find  the  volume  of  the  solid  formed  by  revolving  about  the 
line  x  —  3  the  figure  bounded  by  the  parabola  f  =  8.r  and  the  line 
x  =  2. 

61.  Find  the  volume  of  the  solid  formed  by  revolving  about  the 
line  y  =  —a  the  figure  bounded  by  the  curve  y  =  sinx}  the  lines 

77- 

x  —  0  and  x  =  —  >  and  the  line  y  =  —  a. 

62.  A  right  circular  cone  with  vertical  angle  2a  has  its  vertex  at 
the  center  of  a  sphere  of  radius  a.  Find  the  volume  of  the  portion 
of  the  sphere  intercepted  by  the  cone. 

63.  A  variable  equilateral  triangle  moves  with  its  plane  perpen- 
dicular to  the  axis  of  I/  and  the  ends  of  its  base  respectively  on  the 
parts  of  the  curves  y2=16u.r  and  f  =  4  ax  above  the  axis  of  x. 
Find  the  volume  generated  by  the  triangle  as  it  moves  a  distance  a 
from  the  origin. 

64.  Find  the  volume  of  the  solid  formed  by  revolving  about  OX 

the  plane  figure  bounded  by  the  cissoid  f  — >  the  line  x  =  a, 

and  the  axis  of  x. 

65.  A  right  circular  cylinder  of  radius  a  is  intersected  by  two 
planes,  the  first  of  which  is  perpendicular  to  the  axis  of  the  cylinder, 
and  the  second  of  which  makes  an  angle  6  with  the  first.    Find  the 


290  APPLICATIONS  OF  INTEGRATION 

volume  of  the  portion  of  the  cylinder  included  between  these  two 
planes  if  their  line  of  intersection  is  tangent  to  the  circle  cut  from 
the  cylinder  by  the  first  plane. 

66.  On  the  double  ordinate  of  the  four-cusped  hypocycloid 
x*  +  y*  =  Q>*  as  base  an  isosceles  triangle  is  constructed  with  its 
altitude  equal  to  the  ordinate  and  its  plane  perpendicular  to  the 
plane  of  the  hypocycloid.  Find  the  volume  generated  by  the 
triangle  as  it  moves  from  x  =  —  a  to  x  —  a. 

67.  Find  the  volume  of  the  solid  formed  by  revolving  about  OY 

8<z3 
the  plane  figure  bounded  by  the  witch  y  =  -= : — -  and  the  line  y  =  a. 

68.  Find  the  volume  of  the  solid  formed  by  revolving  about  the 
line  y  =  a  the  plane  figure  bounded  by  the  line  y  =  a  and  the  witch 

8  a* 

69.  Find  the  volume  of  the  solid  bounded  by  the  surface  formed 
by  revolving  the  witch  y  =  — — ^  about  its  asymptote. 

70.  Find  the  volume  of  the  wedge-shaped  solid  cut  from  a  right 
circular  cylinder  of  radius  a  and  altitude  h  by  two  planes  which 
pass  through  a  diameter  of  the  upper  base  and  are  tangent  to  the 
lower  base. 

71.  Two  circular  cylinders  with  the  same  altitude  h  have  the 
upper  base,  of  radius  a,  in  common.  Their  other  bases  are  tangent 
at  the  point  where  the  perpendicular  from  the  center  of  the  upper 
base  meets  the  plane  of  the  lower  bases.  Find  the  volume  common 
to  the  two  cylinders. 

72.  Find  the  volume  of  the  ring  solid  formed  by  revolving  the 
(x  _  ,/)2       ?/a 

ellipse  * ~-  +  j-%  =  1  around  OY {d  >  a). 

73.  The  cap  of  a  stone  post  is  a  solid  of  which  every  horizontal 
cross  section  is  a  square.  The  corners  of  all  the  squares  lie  in  a 
spherical  surface  of  radius  8  in.  with  its  center  4  in.  above  the  plane 
of  the  base.    Find  the  volume  of  the  cap. 

74.  Find  the  volume  of  the  solid  formed  by  revolving  about  the 
line  x  =  —  2  the  plane  figure  bounded  by  that  line,  the  parabola 
y2  =  &x,  and  the  lines  y  =  ±  2. 


PROBLEMS  291 

75.  Find  the  volume  of  the  solid  formed  by  revolving  about  the 
line  x  =  2  the  plane  figure  bounded  by  the  curve  if  =  4  (2  —  x)  and 
the  axis  of  y. 

76.  A  variable  circle  moves  so  that  one  point  is  always  on  OY,  its 

x2      if 
center  is  always  on  the  ellipse  —2  +  'j^  =  1,  and  its  plane  is  always 

perpendicular  to  OY.    Required  the  volume  of  the  solid  generated. 

77.  Find  the  volume  of  the  solid  generated  by  revolving  about 

x3 
the  asymptote  of  the  cissoid  y2  = the  plane  area  bounded  by 

the  curve  and  the  asymptote. 

78.  Find  the  volume  of  the  solid  formed  by  revolving  about 
OX  the  plane  figure  bounded  by  OX  and  an  arch  of  the  cycloid 

x  —  o  ((f)  —  sin  <f>),  y  =  a  (1  —  cos  </>). 

79.  Find  the  volume  of  the  solid  generated  by  revolving  the 
cardioid  /•  =  a(l  +  cos  0)  about  the  initial  line. 

80.  A  cylinder  passes  through  two  great  circles  of  a  sphere  which 
are  at  right  angles  to  each  other.    Find  the  common  volume. 

81.  Find  the  length  of  the  semicubical  parabola  //-  =  (x  —  2)3 
from  its  point  of  intersection  with  the  axis  of  x  to  the  point  (6,  8). 

82.  Find  the  length  of  the  catenary  y  =  „  I '  > "  +  e  " )  from  x  =  0 
to  x  =  It. 

83.  Find    the    total    length    of    the    four-cusped    hypocycloid 

Xs  +  yJ  =  «¥- 

84.  Show  that  the  length  of  the  logarithmic  spiral  r  =  c"e  bel  ween 
any  two  points  is  proportional  to  the  difference  of  the  radius  vectors 
of  the  points. 

Q 

85.  Find  the  complete  length  of  the  curve  r  =  «sin3-- 

86.  Find    the    length  of   the    curve    y  =  a log  —y :  from   the 

a1  —  x2 
.    .  a 

origin  to  *  =  -  • 

a 

87.  Find  the  length  from  cusp  to  cusp  of  the  cycloid 

x  =  a  (<j>  —  sin  <£),     y  =  a  (1  —  cos  <f>). 


292  APPLICATIONS  OF  INTEGRATION 

88.  From  equidistant  points  on  an  arch  of  the  cycloid 

x  =  a  (</>  —  sin  </>),     y  =  a(l  —  cos  <f>), 
perpendiculars  are  drawn  to  the  base  of  the  arch.   What  is  their 
average  length  ? 

89.  From  a  spool  of  thread  2  in.  in  diameter  three  turns  are 
unwound.  If  the  thread  is  held  constantly  tight,  what  is  the  length 
of  the  path  described  by  its  end  ? 

ex  -f-  1 

90.  Find  the   length  of   the  curve  y  =  log from  x  =  1 

to  x  =  2.  e  ~1 

91.  Find  the  mean  distance  of  all  points  on  the  circumference 
of  a  circle  of  radius  a  from  a  given  point  on  the  circumference. 

92.  Find  the  length  of  the  spiral  of  Archimedes,  r  =  a$,  from 
the  pole  to  the  end  of  the  first  revolution. 

93.  Find  the  length  of  the  curve  8a3^/  =  x4  +  6«¥  from  the 
origin  to  the  point  x  =  2  a. 

94.  The  parametric  equations  of  a  curve  are 

x  =  50  (1  -  cos  9)  +  50  (2  -  0)  sin  9,  y  =  50  sin  9  +  50  (2  -  9)  cos  9. 
Find  the  length  of  the  curve  between  the  points  9  =  0  and  0  =  2. 

95.  Find  the  length  of  the  cardioid  r  =  a  (1  +  cos  6). 

96.  Find  the  mean  length  of  the  radius  vectors  drawn  from  the 
pole  to  equidistant  points  of  the  cardioid  r  —  -(1  +  cos  9). 

9 

97.  Find  the  length  of  the  curve  r  =  a  cos5-  from  the  pole  to 

o 

the  point  in  which  the  curve  intersects  the  initial  line. 

98.  Find  the  length  of  the  tractrix  (§  200) 

V  =  T?  log ,  fl  -Va2  -  x2 

from  x  =  h  to  x  =  a. 

99.  Find  the  area  of  a  zone  of  height  h  on  a  sphere  of  radius  a. 

100.  Find  the  area  of  the  surface  formed  by  revolving  about  OX 
the  hypocycloid  x  =  a  cos3  0,  y  =  a  sin3  9. 

101.  Find  the  area  of  the  surface  formed  by  revolving  about  the 
line  x  =  a  the  portion  of  the  hypocycloid  x  =  a  cos3  0,  y  =  a  sin30, 
which  is  at  the  right  of  OY. 


PROBLEMS  293 

102.  Find  the  area  of  the  surface  formed  by  revolving  about  the 

a    -       _- 
tangent  at  its  lowest  point  the  portion  of  the  catenary  i/  —  -(e"-\-e  «) 

between  x  =  —  h  and  x  =  h. 

103.  Find  the  area  of  the  surface  formed  by  revolving  about  the 
initial  line  the  cardioid  r  =  a  (1  +  cos  6). 

104.  Find  the  area  of  the  surface  formed  by  revolving  an  arch  of 
the  cycloid  x  =  a(cf>  —  sin  <f>),  y  =  a(l  —  cos  <f>)  about  the  tangent  at 
its  highest  point. 

105.  Find  the  area  of  the  surface  formed  by  revolving  about  OY 
a  +  vV  —  ./•- 


the  traetrix  (§  200)  //  =  7^  log- —  Va?  —  x*. 

2  a  —  V"-  —  ./•' 

106.  Find  the  area  of  the  surface  formed  by  revolving  the  lem- 
oiscate  r3  =  2  a2  cos  2  6  about  the  initial  line. 

107.  Find  the  area  of  the  surface  formed  by  revolving  the  lem- 
niscate  y3=  2  a2  cos  2  6  about  the  line  9  =  90°. 

108.  A  positive  charge  m  of  electricity  is  fixed  at  0.  The  repul- 
sion on  a  unit  charge  at  a  distance  x  from  (>  is  -■  Find  the  work 
done  in  bringing  a  unit  charge  from  infinity  to  a  distance  a  from  O. 

109.  Assuming  that  the  force  required  to  stretch  a  wire  from  the 

length  a  to  the  length  a  -f-as  is  proportional  to     >  and  that  a  force 

of  lib.  stretches  a  wire  of  36  in.  in  length  to  a  length  .01  in. 
greater,  find  the  work  done  in  stretching  the  wire  from  •">•'>  in. 
to  39  in. 

110.  A  body  moves  in  a  straighl  line  according  to  the  formula 
x  =  cf,  where  x  is  the  distance  traversed  in  the  time  t  II'  the  re- 
sistance of  the  air  is  proportional  to  the  square  of  the  velocity,  find 
the  work  done  against  the  resistance  of  the  air  as  the  body  moves 
from  x  =  0  to  x  =  a. 

ill.  Assuming  that  below  the  surface  of  the  earth  the  force  of 
the  earth's  attraction  varies  directly  as  the  distance  from  the  earth's 
center,  find  the  work  done  in  moving  a  weight  of  w  pounds  from  a 
point  a  miles  below  the  surface  of  the  earth  to  the  surface. 

112.  Assuming  that  above  the  surface  of  the  earth  the  force  of 
the  earth's  attraction  varies  inversely  as  the  square  of  the  distance 


294  APPLICATIONS  OF  INTEGRATION 

from  the  earth's  center,  find  the  work  done  in  moving  a  weight  of 
m  pounds  from  the  surface  of  the  earth  to  a  distance  a  miles  above 
the  surface. 

113.  A  wire  carrying  an  electric  current  of  magnitude  C  is  bent 
into  a  circle  of  radius  a.  The  force  exerted  by  the  current  upon  a 
unit  magnetic  pole  at  a  distance  x  from  the  center  of  the  circle  in 
a  straight  line  perpendicular  to  the  plane  of  the  circle  is  known  to 

be -•    Find  the  work  done  in  bringing  a  unit  magnetic  pole 

from  infinity  to  the  center  of  the  circle  along  the  straight  line  just 
mentioned. 

114.  A  spherical  bag  of  radius  5  in.  contains  gas  at  a  pressure 
equal  to  15  lb.  per  square  inch.  Assuming  that  the  pressure  is  in- 
versely proportional  to  the  volume  occupied  by  the  gas,  find  the 
work  required  to  compress  the  bag  into  a  sphere  of  radius  4  in. 

115.  A  piston  is  free  to  slide  in  a  cylinder  of  cross  section  S. 
The  force  acting  on  the  piston  is  equal  to  pS,  where  p  is  the  pres- 
sure of  the  gas  in  the  cylinder,  and  a  pressure  of  7.7  lb.  per  square 
inch  corresponds  to  a  volume  of  2.5  cu.  in.  Find  the  work  done 
as  the  volume  of  the  cylinder  changes  from  2.5  cu.  in.  to  5  cu.  in., 
(1)  assuming  pv  =  k,  (2)  assuming  pv1A  —  k. 

116.  Find  the  total  pressure  on  a  vertical  rectangle  with  base  8 
and  altitude  12,  submerged  so  that  its  upper  edge  is  parallel  to  the 
surface  of  the  liquid  at  a  distance  5  from  it. 

117.  Find  the  depth  of  the  center  of  pressure  of  the  rectangle 
in  the  previous  problem. 

118.  Find  the  total  pressure  on  a  triangle  of  base  10  and  altitude  4, 
submerged  so  that  the  base  is  horizontal,  the  altitude  vertical,  and 
the  vertex  in  the  surface  of  the  liquid. 

119.  Show  that  the  center  of  pressure  of  the  triangle  of  the  pre- 
vious problem  lies  in  the  median  three  fourths  of  the  distance  from 
the  vertex  to  the  base. 

120.  Find  the  total  pressure  on  a  triangle  with  base  8  and  altitude  6, 
submerged  so  that  the  base  is  horizontal,  the  altitude  vertical,  and  the 
vertex,  which  is  above  the  base,  at  a  distance  3  from  the  surface  of 
the  liquid. 


PROBLEMS  295 

121.  Find  the  depth  of  the  center  of  pressure  of  the  triangle  of 
the  previous  problem. 

122.  The  centerboard  of  a  yacht  is  in  the  form  of  a  trapezoid  in 
which  the  two  parallel  sides  are  3  and  5  ft.  respectively  in  length, 
and  the  side  perpendicular  to  these  two  is  4  ft.  in  length.  Assuming 
that  the  last-named  side  is  parallel  to  the  surface  of  the  water  at  a 
depth  of  1  ft.,  and  that  the  parallel  sides  are  vertical,  rind  the 
pressure  on  the  board.* 

123.  Find  the  moment  of  the  force  which  tends  to  turn  the  center- 
board  of  the  previous  problem  about  the  line  of  intersection  of  the 
plane  of  the  board  with  the  surface  of  the  water. 

124.  A  dam  is  in  the  form  of  a  regular  trapezoid  with  its  two 
horizontal  sides  400  and  100  ft.  respectively,  the  longer  side  being 
at  the  top  and  the  height  20  ft.  Assuming  that  the  water  is  level 
with  the  top  of  the  dam,  find  the  total  pressure. 

125.  Find  the  moment  of  the  force  which  tends  to  overturn  the 
dam  of  the  previous  problem  by  turning  it  on  its  base  line. 

126.  Find  the  total  pressure  on  a  semiellipse  submerged  with  one 
axis  in  the  surface  of  the  liquid  and  the  other  vertical. 

127.  Find  the  depth  of  the  center  of  pressure  of  the  ellipse  of 
the  previous  problem. 

128.  The  gasoline  tank  of  an  automobile  is  in  the  form  of  a 
horizontal  cylinder,  the  ends  of  which  are  plane  ellipses  20  in. 
high  and  10  in.  broad.  Assuming  n-  as  the  weight  of  a  cubic  inch 
of  gasoline,  find  the  pressure  on  one  end  when  the  gasoline  is 
15  in.  deep-. 

129.  A  parabolic  segment  with  base  15  and  altitude  3  is  sub- 
merged so  that  its  base  is  horizontal,  its  axis  vertical,  and  its  vertex 
in  the  surface  of  the  liquid.    Find  the  total  pressure. 

130.  Find  the  depth  of  the  center  of  pressure  of  the  parabolic 
segment  of  the  previous  problem. 

131.  A  circular  water  main  has  a  diameter  of  5  ft.  One  end  is 
closed  by  a  bulkhead  and  the  other  is  connected  with  a  reservoir  in 
which  the  surface  of  the  water  is  20  ft.  above  the  center  of  the 
bulkhead.     Find  the  total  pressure  on  the  bidkhead. 

*The  weight  of  a  cubic  foot  of  water  may  be  taken  as  G21  lb.  =  J.,  ton. 


296  APPLICATIONS  OF  INTE( J  RATION 

132.  A  pond  of  10  ft  depth  is  crossed  by  a  roadway  with  vertical 
sides.  A  culvert,  whose  cross  section  is  in  the  form  of  a  parabolic 
segment  with  horizontal  base  on  a  level  with  the  bottom  of  the  pond, 
runs  under  the  road.  Assuming  that  the  base  of  the  parabolic  seg- 
ment is  6  ft.  and  its  altitude  4  ft.,  find  the  total  pressure  on  the 
bulkhead  which  temporarily  closes  the  culvert. 

133.  Find  the  pressure  on  a  board  whose  boundary  consists 
of  a  straight  line  and  one  arch  of  a  sine  curve,  submerged  so 
that  the  board  is  vertical  and  the  straight  line  is  in  the  surface 
of  the  water. 

134.  Find  the  center  of  gravity  of  the  semicircumference  of  the 
circle  x2  +  if  =  a2  which  is  above  the  axis  of  x. 

135.  Find  the  center  of  gravity  of  the  arc  of  the  four-cusped 
hypocycloid  x*  +  ys  =  a3  which  is  above  the  axis  of  x. 

136.  Find  the  center  of  gravity  of  a  parabolic  segment. 

137.  Find  the  center  of  gravity  of  the  area  of  a  quadrant  of  an 
ellipse. 

138.  Find  the  center  of  gravity  of  a  triangle. 

139.  Find  the  center  of  gravity  of  the  area  bounded  by  the 
semicubical  parabola  ay*  =  xs  and  any  double  ordinate. 

140.  Find  the  center  of  gravity  of  the  area  bounded  by  the  pa- 
rabola x2  -\-  At/  — 16  =  0  and  the  axis  of  x. 

141.  Find  the  center  of  gravity  of  half  a  spherical  solid  of  con- 
stant density. 

142.  Find  the  center  of  gravity  of  the  solid  formed  by  revolving 

aja      v/2 
about  0  Y  the  surface  bounded  by  the  hyperbola  —  —  'j-2  —  1  and  the 

lines  y  =  0  and  y  =  b. 

143.  Find  the  center  of  gravity  of  a  hemisphere. 

144.  Find  the  center  of  gravity  of  the  surface  of  a  right  circular 
cone. 

145.  Find  the  center  of  gravity  of  the  area  bounded  by  the  curve 
y  =  sin  x  and  the  axis  of  x  between  x  =  0  and  x  =  it. 

146.  Find  the  center  of  gravity  of  the  area  between  the  axes  of 
coordinates  and  the  parabola  x1  +  y*  —  a*. 


PROBLEMS  297 

147.  Find  the  center  of  gravity  of  a  uniform  wire  in  the  form  of 
the  catenary  y  =  -  (ea  -f  e~«)  from  x  =  0  to  x  =  a. 

148.  Find  the  center  of  gravity  of  the  solid  formed  by  revolving 
about  OX  the  surface  bounded  by  the  parabola  f  =  &jpx,  the  axis 
of  x,  and  the  line  x  —  a. 

149.  Find  the  center  of  gravity  of  the  plane  area  bounded  by  the 
two  parabolas  f  =  20x  and  x2  =  20 y. 

150.  Find  the  center  of  gravity  of  the  plane  area  bounded  by  the 
parabola  f  =  4x,  the  axis  of  //,  and  the  line  y  =  4. 

151.  Find  the  center  of  gravity  of  the  solid  formed  by  revolving 
about  OY  the  plane  figure  bounded  by  the  parabola  f=Aj,x,  the 
axis  of  y,  and  the  line  y  =  /.'. 

152.  Find  the  center  of  gravity  of  the  surface  of  a  hemisphere 
when  the  density  of  each  point  in  the  surface  varies  as  its  perpen- 
dicular distance  from  the  circular  base  of  the  hemisphere. 

153.  Find  the  center  of  gravity  of  that  part  of  the  plane  surface 
bounded  by  the  four-cusped  hypocycloid  se  =  acos80,  y  =  a sin80, 
which  is  in  the  first  quadrant. 

154.  Find  the  center  of  gravity  of  the  plane  area  bounded  by  the 
ellipse  -5  +  T7,  =  1,  the  circle  x2  +  y*  =  a2,  and  the  n.xis  of  y. 

155.  Find  the  center  of  gravity  of  the  plane  area  common  to  the 
parabola  x2  —  8  y  =  0  and  the  circle  aP+y2—  128  =  0. 

156.  Find  the  center  of  gravity  of  the  plane  surface  bounded  by 
the  first  arch  of  the  cycloid 

x  =  a((f>  —  sin  <£),    y  =  a  (1—  cos  <£), 
and  the  axis  of  x. 

157.  Find  the  center  of  gravity  of  the  arc  of  the  cycloid 

x  =  a(<f>  —  sin  </>),    y  =  a  (1  —  cos  <£), 
between  the  first  two  cusps. 

158.  Find  the  center  of  gravity  of  the  solid  formed  by  rotating 
about  OX  the  parabolic  segment  bounded  by  f  =  4a  and  x  =  It,  if 

the  density  at  any  point  of  the  solid  equals  -• 


298  APPLICATIONS  OF  INTEGRATION 

159.  Find  the  center  of  gravity  of  the  plane  surface  bounded  by 
the  two  circles  xa  -f  if  =  a1 ,  x1  +  //  —  2  ax  =  0,  and  the  axis  of  x. 

160.  Show  that  the  center  of  gravity  of  a  sector  of  a  circle  lies  on 

.    a 

2    Sm2 
the  line  bisecting  the  angle  of  the  sector  at  a  distance  -  a from 

2 
the  vertex,  where  a  is  the  angle  and  a  the  radius  of  the  sector. 

161.  Find  the  center  of  gravity  of  the  solid  generated  by  revolv- 
ing about  the  line  x  =  a  the  area  bounded  by  that  line,  the  axis  of  x, 
and  the  parabola  if  =  4=px. 

162.  Find  the  center  of  gravity  of  the  plane  area  bounded  by  the 
two  parabolas  x2  —  4tp  (y  —  b)  =  0,  x2  —  kpy  =  0,  the  axis  of  y,  and 
the  line  x  =  a. 

163.  Find  the  center  of  gravity  of  the  arc  of  the  curve  9  ay2  — 
x (x  —  3 a)'2  =  0  between  the  ordinates  x  =  0  and  x  =  3a. 

164.  The  density  at  any  point  of  a  lamina  in  the  form  of  a  para- 
bolic segment  of  height  8  ft.  and  base  6  ft.  is  directly  proportional 
to  its  distance  from  the  base.    Find  the  center  of  gravity. 

165.  Find  the  center  of  gravity  of  the  portion  of  a  spherical 
surface  bounded  by  two  parallel  planes  at  distances  h^  and  h2 
respectively  from  the  center. 

166.  Find  the  center  of  gravity  of  the  solid  formed  by  revolving 
about  OY  the  plane  area  bounded  by  the  parabola  x2  =  4^y  and  any 
straight  line  through  the  vertex. 

167.  Find  the  center  of  gravity  of  the  surface  generated  by  the 
revolution  about  the  initial  line  of  one  of  the  loops  of  the  lemniscate 
r2=2«2cos2  0. 

168.  Prove  that  the  total  pressure  on  a  plane  surface  perpendic- 
ular to  the  surface  of  a  liquid  is  equal  to  the  pressure  at  the  center 
of  gravity  multiplied  by  the  area  of  the  surface. 

169.  Prove  that  the  area  generated  by  revolving  a  plane  curve 
about  an  axis  in  its  plane  is  equal  to  the  length  of  the  curve  multi- 
plied by  the  circumference  of  the  circle  described  by  its  center  of 
gravity. 

170.  Prove  that  the  volume  generated  by  revolving  a  plane  figure 
about  an  axis  in  its  plane  is  equal  to  the  area  of  the  figure  multiplied 
by  the  circumference  of  the  circle  described  by  its  center  of  gravity. 


PEOBLEMS  299 

171.  Find  the  attraction  of  a  uniform  straight  wire  of  length  20 
and  mass  M  upon  a  particle  of  unit  mass  situated  in  the  line  of 
direction  of  the  wire  at  a  distance  3  from  one  end. 

172.  Find  the  attraction  of  a  rod  of  mass  M  and  length  /,  whose 
density  varies  as  the  distance  from  one  end,  on  a  particle  of  unit 
mass  in  its  own  line  and  distant  a  units  from  that  end. 

173.  A  particle  of  unit  mass  is  situated  at  a  perpendicular  dis- 
tance 5  from  the  center  of  a  straight  homogeneous  wire  of  mass  .1/ 
and  length  12.    Find  the  force  of  attraction  of  the  wire. 

174.  Find  the  attraction  due  to  a  straight  wire  of  length  2  /  on  a 
particle  of  unit  mass  lying  on  the  perpendicular  at  the  middle  point 
of  the  wire  and  distant  c  units  from  the  wire,  the  density  of  the  wire 
varying  directly  as  the  distance  from  its  middle  point. 

175.  Find  the  attraction  of  a  homogeneous  straight  wire  of  neg- 
ligible thickness  and  infinite  length  on  a  particle  of  unit  mass  at  a 
perpendicular  distance  c  from  the  central  point  of  the  wire. 

176.  Find  the  attraction  of  a  uniform  wire  of  mass  .1/  bent  into 

77- 

an  arc  of  a  circle  with  radius  5  and  angle  —  upon  a  particle  of  unit 
mass  at  the  center  of  the  circle. 

177.  Find  the  attraction  of  a  uniform  circular  ring  of  radius  a 
and  mass  M  upon  a  particle  of  unit  mass  situated  at  a  distance  c 
from  the  center  of  the  ring  in  a  straight  line  perpendicular  to  the 
plane  of  the  ring. 

178.  Find  the  attraction  of  a  uniform  circular  disk  of  radius  a 
and  mass  M  upon  a  particle  of  unit  mass  situated  at  a  perpendicular 
distance  c  from  the  center  of  the  disk.  (Divide  the  disk  into  con- 
centric rings  and  use  the  result  of  problem  177.) 

179.  Find  the  attraction  of  a  uniform  right  circular  cylinder  with 
mass  M,  radius  of  its  base  a,  and  length  I  upon  a  particle  of  unit 
mass  situated  in  the  axis  of  the  cylinder  produced,  at  a  distance  c 
from  one  end.  (Divide  the  cylinder  into  parallel  disks  and  use  the 
result  of  problem  178.) 

180.  Find  the  attraction  of  a  uniform  straight  wire  of  length  5 
and  mass  M  upon  a  particle  of  unit  mass  situated  at  a  perpendicular 
distance  12  from  the  wire  and  so  that  lines  drawn  from  the  particle 

7T 

to  the  ends  of  the  wire  inclose  an  angle  — • 

o 


CHAPTER  XIV 
SPACE  GEOMETRY 

139.  Functions  of  more  than  one  variable.  A  quantity  z  is 
mid  to  be  a  function  of  two  variables,  x  and  y,  if  the  values  of  z 
are  determined  when  the  values  of  x  and  y  are  given.  This  rela- 
tion is  expressed  by  the  symbols  z  —f(x,  y),  z  =  F(x,  y),  etc. 

Similarly,  u  is  a  function  of  three  variables,  x,  y,  and  z,  if  the 
values  of  u  are  determined  when  the  values  of  x,  y,  and  z  are 
given.  This  relation  is  expressed  by  the  symbols  u  =f(x,  y,  z), 
u  =  F(x,  y,  z),  etc. 

Ex.  1.  If  r  is  the  radius  of  the  base  of  a  circular  cone,  h  its  altitude, 
and  v  its  volume,  v  =  ^  7rr2h,  and  v  is  a  function  of  the  two  variables 
r  and  h. 

Ex.  2.    If  /  denotes  the  centrifugal  force  of  a  mass  m  revolving  with  a 

velocity  v  in  a  circle  of  radius  >',f= ,  and  f  is  a  function  of  the  three 

variables  m,  v,  and  r. 

Ex.  3.    Let  v  denote  a  volume  of  a  perfect  gas,  t  its  absolute  temperature, 

and  p  its  pressure.   Then  —  =  k,  where  k  is  a  constant.   This  equation  may 

be  written  in  three  equivalent  forms:   p  =  k-,  v  —  k-,  t=  -pv,  by  which 

each  of  the  quantities  p,  v,  and  /  is  explicitly  expressed  as  a  function  of 
the  other  two. 

A  function  of  a  single  variable  is  defined  explicitly  by  the 
equation  y  =f(x),  and  implicitly  by  the  equation  F(x,  y)  =  0 
(§  86).  In  either  case  the  relation  between  x  and  y  is  repre- 
sented graphically  by  a  plane  curve.  Similarly,  a  function  of 
two  variables  may  be  defined  explicitly  by  the  equation 
z  =f(x,  y~),  or  implicitly  by  the  equation  F(x,  y,  z)  —  0.  In 
either  case  the  graphical  representation  of  the  function  of  two 
variables  is  the  same,  and  may  be  made  by  introducing  the 
conception  of  space  coordinates. 

300 


RECTANGULAR  COORDINATES  301 

140.  Rectangular  coordinates.  To  locate  a  point  in  space  of 
three  dimensions,  we  may  assume  three  number  scales,  OX,  OY, 
OZ  (rig.  186),  mutually  perpendicular,  and  having  their  zero 
points  coincident  at  0.  They  will  determine  three  planes, 
XO  Y,  YOZ,  ZOX,  each  of  which  is  perpendicular  to  the  other 
two.  The  planes  are  called  the  coordinate  planes,  and  the  three 
lines  OX,  OY,  and  OZ  are  called  the  axes  of  x,  y,  and  z 
respectively,  or  the  coordinate  axes,  and  the  point  0  is  called 
the  origin  of  coordinates. 

Let  P  be  any  point  in  space,  and  through  P  pass  planes 
perpendicular  respectively  to  OX,  0  )',  and  OZ,  intersecting 
them  at  the  points  L,  M,  and  N  respectively.  Then  it'  we 
place  x  —  OL,  y  =  0M,  and  z  =  ON,  it  is  evident  that  to  any 
point  there  corresponds  one,  and  only 
one,  set  of  values   of   x,   y,   and   z ;    and 

that    to    any    set    of   values    of    ./-,    //,    and  n\ jj, 

z    there    corresponds    one,    and    only    one,  s'/—  L— -{/ 

point.     These   values   of   x,   y,    and    z   are 

called  the  coordinates  of  the  point,  which <)'</'_ \l    x 

is  expressed  as  /'(./,  //,  z).  / 

From    the    definition    of    x    it    follows      y  M       -17' 
that  x  is  equal,  in  magnitude  and  direc-  Fi(.    ]S(. 

tion,  to  the  distance  of  the  point  from  the 
coordinate  plane  YOZ.  Similar  meanings  are  evident  for  y 
and  z.  It  follows  that  a  point  may  be  [(lotted  in  several  dif- 
ferent ways  by  constructing  in  succession  any  three  nonparallel 
edges  of  the  parallelepiped  (fig.  18G)  beginning  at  the  origin 
and  ending  at  the  point. 

In  case  the  axes  are  not  mutually  perpendicular,  we  have  a  system  of 
oblique  coordinates.  In  this  case  the  planes  are  passed  through  the  point 
parallel  to  the  coordinate  planes.  Then  x  gives  the  distance  and  the  direc- 
tion from  the  plane  YOZ  to  the  point,  measured  parallel  to  OX,  and  similar 
meanings  are  assigned  to  y  and  z.  It  follows  that  rectangular  coordinates 
are  a  special  case  of  oblique  coordinates. 

141.  Graphical  representation  of  a  function  of  two  variables. 
Let  f(x,  y)  be  any  function  of  two  variables,  and  place 

s=/0,y>  (i) 


302  tiVACK  GEOMETRY 

Then  the  locus  of  all  points  the  coordinates  of  which  satisfy  (1) 
is  -the  graphical  representation  of  the  function  f(sc,  y).  To  con- 
struct this  locus  we  may  assign  values  to  x  and  y,  as  x  —  xx  and 
y  =  yl,  and  compute  from  (1)  the  corresponding  values  of  z. 
There  will  be,  in  general,  distinct  values  of  z,  and  if  (1)  defines 
an  algebraic  function,  their  number  will  be  finite.  The  corre- 
sponding points  all  lie  on  a  line  parallel  to  OZ  and  intersecting 
XOY  at  the  point  Pl(xl,  y^),  and  these  points  alone  of  this  line 
are  points  of  the  locus,  and  the  portions  of  the  line  between 
them  do  not  belong  to  the  locus.  As  different  values  are 
assigned  to  x  and  y,  new  lines  parallel  to  OZ  are  drawn  on 
which  there  are,  in  general,  isolated  points  of  the  locus.  It 
follows  that  the  locus  has  extension  in  only  two  dimensions, 
i.e.  has  no  thickness,  and  is,  accordingly,  a  surface.  Therefore 
the  graphical  representation  of  a  function  of  two  variables  is  a 
surface* 

If  fQx,  y~)  is  indeterminate  for  particular  values  of  x  and  y, 
the  corresponding  line  parallel  to  OZ  lies  entirely  on  the  locus. 

Since  the  equations  z  —f(x,  y)  and  F(x,  y,  z~)  =  0  are  equiv- 
alent, and  their  graphical  representations  are  the  same,  it  follows 
that  the  locus  of  any  single  equation  in  x,  y,  and  z  is  a  surface. 

There  are  apparent  exceptions  to  the  above  theorem  if  we  demand  that 
the  surface  shall  have  real  existence.    Thus,  for  example, 


+  f  + 


is  satisfied  by  no  real  values  of  the  coordinates.    It  is  convenient  in  such 
cases    however,  to  speak  of  "  imaginary  surfaces." 

Moreover,  it  may  happen  that  the  real  coordinates  which  satisfy  the 
equation  give  points  which  lie  upon  a  certain  line,  or  are  even  isolated 
points.    For  example,  the  equation 

x2  +  y-  =  0 

is  satisfied  in  real  coordinates  only  by  the  points  (0,  0,  z)  which  lie  upon 
the  axis  of  z ;  while  the  equation 

x2  +  if  +  z2  =  0 

*  It  is  to  be  noted  that  this  method  of  graphically  representing  a  function 
cannot  be  extended  to  functions  of  more  than  two  variables,  since  we  have  but 
three  dimensions  in  space. 


CYLINDERS  303 

is  satisfied,  as  far  as  real  points  go,  only  by  (0,  0,  0).  In  such  cases  it 
is  still  convenient  to  speak  of  a  surface  as  represented  by  the  equation, 
and  to  consider  the  part  which  may  be  actually  constructed  as  the  real 
part  of  that  surface.  The  imaginary  part  is  considered  as  made  up  of 
the  points  corresponding  to  sets  of  complex  values  of  x,  y,  and  z  which 
satisfy  the  equation. 

142.  Cylinders.  If  a  given  equation  is  of  the  form  F(x,  y)  =  0, 
involving  only  two  of  the  coordinates,  it  might  appear  to  rep- 
resent a  curve  lying  in  the  plane  of  those  coordinates.  But  if 
we  are  dealing  with  space  of  three  dimensions,  such  an  inter- 
pretation would  be  incorrect,  in  that  it  amounts  to  restricting 
z  to  the  value  s  =  0,  whereas,  in  fact,  the  value  of  z  corre- 
sponding to  any  simultaneous  values  of  x  and  y  satisfying 
the  equation  F(x,  y)=0  may  be  anything  whatever.  Hence, 
corresponding  to  every  point  of  the  curve  F(x,  //)  =  0  in  the 
plane  XOY,  there  is  an  entire  straight  line,  parallel  to  OZy 
on  the  surface  F(.r,  y)  =  0.  Such  a  surface  is  a  cylinder, 
its  directrix  being  the  plane  curve  /''(./•,  y)  =  Q  in  the  plane 
2  =  0,  and  its  elements  being  parallel  to  OZ,  the  axis  of  the 
coordinate  not  present. 

For  example,  .r  +  //"  =  <c  is  the  equation  of  a  circular  cylinder, 
its  elements  being  parallel  to  OZ,  and  its  directrix  being  the  circle 
a?+y*=  a2  in  the  plane  XOY. 

In  like  manner  sa=  ky  is  the  equation  of  a  parabolic  cylinder, 
its  elements  being  parallel  to  OX,  and  its  directrix  being  the 
parabola  z'2=ky  in  the  plane  ZOY. 

If  only  one  coordinate  is  present  in  the  equation,  the  locus  is  a 
number  of  planes.  For  example,  the  equation  x'1  —  ( (t  +  t>)x+ab=0 
may  be  written  in  the  form  (x—  a)  (x—  l>)  =  0,  which  represents 
the  two  planes  x  —  a  =  0  and  x  —  b  —  0.  Similarly,  any  equation 
involving  only  one  coordinate  determines  values  of  that  coordinate 
only  and  the  locus  is  a  number  of  planes. 

Regarding  a  plane  as  a  cylinder  of  which  the  directrix  is  a 
straight  line,  we  may  say  that  any  equation  not  containing  all 
the  coordinates  represents  a  cylinder. 

If  the  axes  are  oblique,  the  elements  of  the  cylinders  are  not 
perpendicular  to  the  plane  of  the  directrix. 


304 


SPACE  GEOMETRY 


143.  Other  surfaces.  The  surface  represented  by  any  equa- 
tion /'(./-,  //,  z)  =  0  may  be  studied  by  means  of  sections  made 
by  planes  parallel  to  the  coordinate  planes.  If,  for  example, 
we  places  =  <>  in  the  equation  of  any  surface,  the  resulting 
equation  in  x  and  y  is  evidently  the  equation  of  the  plane 
curve  cut  from  the  surface  by  the  plane  XOY.  Again,  if  we 
place  2  =  21?  where  zx  is  some  fixed  finite  value,  the  resulting 
equation  in  x  and  y  is  the  equation  of  the  plane  curve  cut 
from  the  surface  by  a  plane  parallel  to  the  plane  XOY  and  zl 
units  distant  from  it,  and  referred  to  new  axes  O'X'  and  O'Y', 
which  are  the  intersections  of  the  plane  z  =  zx  with  the  planes 
XOZ  and  YOZ  respectively;  for  by  placing  z  =  zx  instead  of 
2=0,  we  have  virtually  transferred  the  plane  XOY,  parallel  to 
itself,  through  the  distance  z^ 

In  applying  this  method  it  is  advisable  to  find  first  the  three 
plane  sections  made  by  the  coordinate  planes  x  =  0,  i/  =  0,  z  =  Q. 
These  alone  will  sometimes  give  a  general  idea  of  the  appearance 
of  the  surface,  but  it  is  usually  desirable  to  study  other  plane  sec- 
tions on  account  of  the  additional  information  that  may  be  derived. 

The  following  surfaces  have  been  chosen  for  illustration  because 
it  is  important  that  the  student  should  be  familiar  Avith  them. 

Ex.  1.    Ax  +  By  +  Cz  +  D  =  0. 

Placing-  z  =  0,  we  have  (fig.  187) 

Ax  +  By  +  D  =  0.  (1) 

Hence  the  plane  XOY  cuts  this  surface 
in  a  straight  line.  Placing  y  =  0  and  then 
x  =  0,  we  find  the  sections  of  this  surface 
made  by  the  planes  ZOX  and  YOZ  to  be 
respectively  the  straight  lines 

Ax  +  Cz  +  D=0,  (2) 

and  By  +  Cz  +  D=0.  (3) 

Placing  z  =  zv  we  have  A  x  +  By  +  ( 'z1  +  D  =  0,  (4) 

which  is  the  equation  of  a  straight  line  in  the  plane  z  —  zv  The  line  (1) 
is  parallel  to  the  line  (1),  since  they  make  the  angle  tan-1/  —  —J  with  the 
parallel  lines  O'X'  and  OX  and  lie  in  parallel  planes.    To  find  the  point 


SURFACES 


305 


where  (4)  intersects  the  plane  XOZ,  we  place  y  =  0,  and  the  result 
Ax  +  Cz1  +  D=  0  shows  that  this  point  is  a  point  of  the  line  ("2).  This 
result  is  true  for  all  values  of  zr  Hence  this  surface  is  the  locus  of  a 
straight  line  which  moves  along  a  fixed  straight  line  always  remaining 
parallel  to  a  given  initial  position;  hence  it  is  a,  plane. 

Since  the  equation  Ax  +  By  +  Cz  +  D=  0  is  the  most  general  equa- 
tion of  the  first  degree  in  the  three  coordinates,  we  have  proved  that 
the  locus  of  every  lunar  equation  in  rectangular  spaa    coordinates  is  a  plane. 


+  !>//-,  where   a  >  0 


Ex.  2.    s 
&>0. 

Placing  2  =  0,  we  have 

a./2  +  by*  =  0, 

and  hence  the  XOY  plane 

surface  in  a  point  (fig.  L88) 

,j  =  (),  we  have 

z  =  il.i-, 

which  is  the  equation  of  a 

with  its  vertex  at  0  ami  its  axis  along 

OZ.    Placing  x  =  <•,  we  have 

z  =  /,n  (3) 

which  is  also  the  equation  of  a  pa- 
rabola with  its  vertex  at  0  and  its 
axis   along   (>Z. 

Placing  z  =  z1,  where  zx  >  0,  we 
may  write  the  resulting  equation  in 
t  he  form 


b 


(1) 


and 


As    the 


-1  \     -1 

which  is  the  equation  of  an  ellipse  with  semiaxes 

\  (I  \   h 

plane  recedes  from  the  origin,  i.e.  as  -n  increases,  it   is  evident  thai  the 

ellipse  increases  in  magnitude.   It  is  also  evident  thai  the  ends  of  the  axes 

of  the  ellipse  lie  on  the  parabolas  (•_')  and  (:">). 

If  we  place  z  =  —  zv  the  result  may  he  w  litten  in  the  form 


— y 


l, 


and  hence  there  is  no  part  of  this  surface  on  the  negative  side  of  the 
plane  XOY. 

The  surface  is  called  an  elliptic  paraboloid,  and  evidently  may  l>e  gener- 
ated by  moving  an  ellipse  of  variable  magnitude  always  parallel  to  the 
plane  XOY,  the  ends  of  its  axes  always  lying  respectively  on  the  parabolas 
2  =  <u'2  and  z  =  by2. 


306 


SPACE  GEOMETKY 


Ex.  3. 


Placing: 


J  f 

i2      b* 

c2 

1. 

2  =  0,  we  have 

a2      62 

=  1, 

(1) 


which  is  the  equation  of  an  ellipse 
with  semiaxes  a  and  b  (fig.  189). 
Placing  y  —  0,  we  have 


=  1. 


(2) 


Fig.  189 


which  is  the  equation  of  an 
hyperbola  with  its  transverse 
axis  along  OX  and  its  conjugate  axis  along  OZ.    Placing  x  =  0,  we  have 


=  1, 


(3) 


which  is  the  equation  of  an  hyperbola  with  its  transverse  axis  along  OY 
and  its  conjugate  axis  along  OZ. 

If  we  place  z  =  ±  zv  and  write  the  resulting  equation  in  the  form 


°!H)+*(1+S) 


(i) 


we  see  that  the  section  is  an  ellipse  with  semiaxes 


n\i+i* 


and  h  \ /l  +  ZL, 

which  accordingly  increases  in  magnitude  as  the  cutting  plane  recedes  from 
the  origin,  and  that  the  surface  is  symmetrical  with  respect  to  the  plane 
XOY,  the  result  being  independent  of  the  sign  of  z1. 

Accordingly  this  surface,  called  the  imparted  hyperboloid  or  the  hi/per- 
boloid  of  one  sheet,  may  be  generated  by  an  ellipse  of  variable  magnitude 
moving  always  parallel  to  the  plane 
XOY  and  with  the  ends  of  its  axes 
always   lying    on   the   hyperbolas 

x1      z2  ?/2      ~2 

%■  -  z-  =  1  and  %-  -  z-  =  1. 


Ex.  4.    -2  +  f-  - 1-  =  0 

<r       h-       c2 


This  surface  (fig.  190)  is  a  cone,  with 
OZ  as  its  axis  and  its  vertex  at  0. 


Fig.  190 


SURFACES 


307 


T*  II*  ~a 

Ex.  5.    -a  +  g  +  5  =  1- 

a-       6-       c- 

This .  surface   (fig.  191)   is   the 
ellipsoid. 


Ex.  6. 


/;- 


Fig.  191 


This  surface  (fig.  192)  is  the 
biparled  hyperboloid  or  the  hyper- 
boloid  of  two  sheets. 

The  discussions  of  the  last  three  surfaces  are  very  similar  to  that  of 
the    imparted    hyperboloid,    and     for 
that    reason    they   have    been   left  to 
the   student. 

Ex.  7.  ~  =  ax2  —  by2,  where  a  >  0, 
I  >  0. 

Placing  z  =  0,  we  obtain  the  equa- 
tion 


ax2  -  by2  =  0, 


(1) 


i.e.    two    straight    lines    intersecting 
at    the     origin    (fig.   193).      Placing 

y  =  0,  we  have 


(l.r- 


(2) 


Fig.  192 


the  equation  of  a   parabola  with   its 

v^i^x    at    0    and    its    axis    along    the    positive    direction    of    OZ. 

IS  x  =  0,  we  have 

=  -  by*,      (3) 

the  equation  of  a  parabola 

with  its  vertex  at  0  and 
its  axis  along  the -negat  ive 
direction  of  OZ. 

Placing    ./•  =  ±  xv    we 
have 

Z  =  (,.r{  -  /;//'-, 


or  \f 


±(z-ax?),  (1) 


Fig.  193 


a  parabola  with  its  axis 
parallel  to  OZ  and  its 
vertex   at  a  distance   ax* 

from  the  plane  XOY.     It  is  evident,  moreover,  that  the   surface  is  sym- 
metrical   with    respect    to    the    plane    YOZ,    and    that    the    vertices    of 


308 


SPACE  GEOMETKY 


these  parabolas,  as  different  values  are  assigned  to  xv  all  lie  on  the 
parabola  z  =  ax2. 

Hence  this  surface  may  be  generated  by  the  parabola  z  =  —  by1  moving 
always  parallel  to  the  plane  YOZ,  its  vertex  lying  on  the  parabola  z  =  ax". 
The  surface  is  called  the  hyperbolic  paraboloid. 

The  reason  for  the  name  given  to  this  surface  becomes  more  evident  if 
two  more  sections  are  made. 

Placing  z  =  zv  where  z1  >  0,  we  have  zx  =  ax2  —  hy2,  or 

f-Lx2-Lf-  =  l,  (5) 


an  hyperbola  with  its  transverse  axis  parallel  to   OX,  the  ends  of  the 
transverse  axis  lying  on  the  parabola  z  =  ax'2. 

If  z  =  —  zv  we  may  write  the  equation  in  the  form 


■v- 


(6) 


an  hyperbola  with  its  transverse  axis  parallel  to   OY,  the  ends  of  the 
transverse  axis  lying  on  the  parabola  z  =  —  by'1. 


Ex.  8.  z  =  kxy,  where 
fc>0. 

This  surface  is  a 
special  case  of  the  hyper- 
bolic paraboloid  of  Ex.  7, 
in  which  b=a.  The  proof 
of  this  statement  is  as 
follows : 

If  h  =  a,  the  equation 
of  the  surface  of  Ex.  7  is 

z  =  a(x*-y*).    (1) 


Fig.  194 


Revolve     the     planes 
XOZ    and    YOZ    about 

the  axis  OZ,  which  is  held  fixed,  through  an  angle  of  —  45°  into  new 
positions  X'OZ  and  Y'OZ.    By  §  19,  the  formulas  of  transformation  are 

,  _  x'  +  y'  _  —  x  +  // 

Z-      ,  X-  , 


V2 


Substituting  these  values  in  (1),  and  simplifying,  we  have 

z'  =  2  a  x'y', 
which  is  the  equation  given  above  with  k  —  2a. 


(2) 


SURFACES  309 

The  discussion  of  the  plane  sections  of  the  surface  (fig.  194)  made  by 
the  planes  parallel  to  the  coordinate  planes  is  left  to  the  student. 

If  b  ^  a,  we  can  make  a  similar  transformation  by  using  the  formulas 
of  §  21,  and  the  result  will  be  ~'  =  k/y,  only  the  coordinates  will  not  be 
rectangular. 

144.  Surfaces  of  revolution.  If  the  sections  of  a  surface 
made  by  planes  parallel  to  one  of  the  coordinate  planes  are 
circles  with  their  centers  on  the  axis  of  coordinates  which  is 
perpendicular  to  the  cutting  planes,  the  surface  is  a  surface  of 
revolution  (§  133)  with  that  coordinate  axis  as  the  axis  of 
revolution.  This  will  always  occur  when  the  equation  of  the 
surface  is  in  the  form  F{\f.r  +  >/%  z)  =  0,  which  means  that  the 
two  coordinates  x  and  y  enter  only  in  the  combination  v.r  + //- ; 
for  if  we  place  z  =  z1  in  this  equation  to  find  the  corresponding 
section,  and  solve  the  resulting  equation  for  ./•"  +  //",  we  have, 
as  a  result,  the  equation  of  one  or  more  circles,  according  to 
the  number  of  roots  of  the  equation  in  x1  +  y'1. 

Again,  if  we  place  x  =  0,  we  have  the  equation  /•'(//,  z)  =  0, 
which  is  the  equation  of  the  generating  curve  in  the  plane 
YOZ.  Similarly,  if  we  place  //  =  0,  we  have  /•'(./-,  g)  =  0,  which 
is  the  equation  of  the  generating  curve  in  the  plain'  XOZ.  It 
should  be  noted  that  the  coordinate  which  appears  uniquely  in 
the  equation  shows  which  axis  of  coordinates  is  the  axis  of 
revolution. 

Ex.  1.    Show  that  the  imparted  hvperboloid  ',  —  •-  +  -=  1  is  a  surface 

"'"      //_      "'" 
of  revolution. 

Writing  this  equation  in  the  form 

t±*  -f-l  =  Qt 

a-  b- 

we  see  that  it  is  a  surface  of  revolution  with  OY  as  the  axis. 

c2      >r 
Placing  z  =  0,  we  have  —  —  2-=l,  an   hyperbola,  as  the  generating 
a-      b- 

curve.    The  hyperbola  was  revolved  about  its  conjugate  axis. 

Conversely,  if  we  have  any  plane  curve  F(x,  z)  =  0  in  the 
plane  XOZ,  the  equation  of  the  surface  formed  by  revolving  it 
about  OZ  as  an  axis  is  F(Vx2  +  y2,  z)  =  0,  which  is  formed  by 
simply  replacing  x  in  the  equation  of  the  curve  by  Vx2  +  y1. 


310 


SPACE  GEOMETRY 


Ex.  2.    Find  the  equation  of  the  sphere  formed  by  revolving  the  circle 
+  z2  =  a2  about  OX  as  an  axis. 


Replacing  a  by  Vy2  +  z2,  we  have  as  the  equation  of  the  sphere, 

x2  +  y-  +  z2  =  a2. 

This  equation  may  also  be 
found  directly  from  a  figure. 

Let  Px  (fig.  195)  be  any  point 
of  the  circle,  and  .let  P  be  any 
point  of  the  sphere,  on  the  circle 
described  by  Pv  Since  Px  is  a 
point  of  the  circle, 

OZ2  +  TP{  =  a2.         (1) 

But  LP^LP=VlM2+Mp'2- 
Substituting  this  value  of  LP1 
in  (1),  we  have 

OZ2  +  LM2  +  MP2  =  a2, 

or  x2  +  f  +  z2  =  a2, 

as  the  equation  of  the  sphere. 

145.  Projection.  The  projection  of  a  point  on  a  straight  line 
is  defined  as  the  point  of  intersection  of  the  line  and  a  plane 
through  the  point  perpendicular  to  the  line.  Hence,  in  fig.  186, 
L,  M,  and  N  are  the  projections  of  the  point  P  on  the  axes 
of  x,  y,  and  z  respectively. 

The  projection  of  one  straight 
line  of  finite  length  upon  a  second 
straight  line  is  the  part  of  the 
second  line  included  between  the 
projections  of  the  ends  of  the  first 
line,  its  direction  being  from  the 
projection  of  the  initial  point  of 
the  first  line  to  the  projection  of 
the  terminal  point  of  the  first 
line.  In  fig.  196,  for  example, 
the  projections  of  A  and  B  on  MN  being  A'  and  B'  respectively, 
the  projection  of  AB  on  MJSF  is  A'B',  and  the  projection  of  BA 
on  MN  is  B'A'.  If  MN  and  AB  denote  the  positive  directions 
respectively  of  these  lines,  it  follows  that  A'B'  is  positive  when 


M- 


Q 

4--- 

F^ 

A' 

1 

t 

B' 

Fig.  196 


PROJECTION 


311 


it  has  the  same  direction  as  MNy  and  is  negative  when  it  has 
the  opposite  direction  to  MN. 

In  particular,  the  projection  on  OX  of  the  straight  line  i^ 
drawn  from  Pl(xv  yv  zj  to  P2(x2,  y.v  z2)  is  LXLV  where  OLx  =  x 
and  0L2=  x2.  But  LxL2=  x2  —  xv  by  §  3.  Hence  the  projection 
of  PXP2  on  OX  is  x2  —  xx ;  and,  similarly,  its  projections  on  0  Y 
and  OZ  are  respectively  y2— yt  and  z2  —  zx. 

If  we  define  the  angle  between  any  two  lines  in  space  as  the 
angle  between  lines  parallel  to  them  and  drawn  from  a  common 
point,  then  the  projection  of  one  *tr<ii:ilit  line  <>n  a  second  is  the 
product  of  the  length  of  the  first  line  and  the  cosine  of  the  angle 
between  the  positive  directions  of  the  two  lines.  Then,  if  $  is  the 
angle  between  AB  and  MN  (fig.  196), 

A'B'  =  ^17?  cos  4>. 

To  prove  this  proposition,  draw  A'C  parallel  to  .//;  and  meet- 
ing the  plane  ST  at  C.  Then  A'C  =  AB,  and  A'B' =  A'C  cos  <f>, 
by  §  2,  whence  the  truth 
of  the  proposition  is 
evident. 

Defining  the  projec- 
tion of  a  broken  line 
upon  a  straight  line  as 
the  sum  of  the  projec- 
tions of  its  segments,  we 
may  prove,  as  in  §  2, 
that  the  projections  on 
any  straight  line  of  a 
broken  line  and  the  straight  line  joining  its  ends  are  the  same. 

We  will  now  show  that  the  projection  of  a)iy  plane  area  )i/»>n 
another  plane  is  the  product  of  that  area  and  the  cosine  of  the 
angle  between  the  planes. 

Let  X'OY  (fig.  197)  be  any  plane  through  OY  making  an 
angle  <f>  with  the  plane  XOY.  Let  A'B'  be  any  area  in  X'OY 
such  that  any  straight  line  parallel  to  OX'  intersects  its 
boundary  in  not  more  than  two  points,  and  let  AB  be  its 
projection   on  XOY. 


312  SPACE  GEOMETRY 

Then  (§  125)  area  A'B'  =  C (x[  -  x[)  dy,  (1) 

the   limits    of    integration    being    taken    so    as   to    include   the 
whole  area. 

In  like  manner,        area AB  —  I  (x2  —  xi)dy,  (2) 

the  limits  of  integration  being  taken  so  as  to  cover  the  whole  area. 
But  the  values  of  y  are  the  same  in  both  planes,  since  they  are 
measured  parallel  to  the  line  of  intersection  of  the  two  planes; 
and  hence  the  limits  in  (1)  and  (2)  are  the  same.  Since  the  x 
coordinate  is  measured  perpendicular  to  the  line  of  intersection, 
x2  =  x'2  cos  <£,  x1  =  x[  cos  <f>,  and  (2)  becomes 


area  AB 


=  /  (./•.(  —  x[)  cos  <j>  dy 
=  cos  <f>  j  (x'2  —  .Tj')  dy 
=  (cos  <£)  (area  A'B'). 


H>-: 


146.  Components  of  a  directed  straight  line.  Let  BXB2  (fig.  198) 
be  a  straight  line,  the  direction  of  which  is  from  Bx  to  B,.  Through 
Bx  and  B2  pass  planes  parallel  respectively 
to  the  coordinate  planes,  thereby  forming 
on  BJ!2  as  a  diagonal  a  rectangular  paral- 
lelepiped with  its  edges  parallel  to  the 
coordinate  axes.  The  lines  BtQ,  BXB,  and 
BltS,  considered  with  respect  to  both  length 
and  direction,  are  called  the  components  of 
7^.  It  is  evident  that  they  are  the  projec- 
tions of  i^on  OX,  OY,  and  OZ  respectively. 

Conversely,  the  components  of  a  straight  line  will  determine 
its  direction  and  length,  but  not  its  position ;  for  if  the  compo- 
nents are  given  equal  to  a,  b,  and  e,  we  may  lay  off,  from  any 
point  Ij,  a  straight  line  parallel  to  OX  and  equal  to  a  in  length, 
a  straight  line  parallel  to  OY  and  equal  to  b  hi  length,  and  a 
straight  line  parallel  to  OZ  and  equal  to  c  in  length.  These 
three  lines  determine  the  edges  of  a  rectangular  parallelepiped, 
and  hence  determine  the  diagonal  drawn  from  Bv  That  is,  if  BXQ 
(fig.  198)  is  laid  off  equal  to  a,  B^  equal  to  b,  and  BXS  equal 


DISTANCE  313 

to  c,  the  rectangular  parallelepiped  is  determined,  and  hence  the 
diagonal  PXP2  is  determined  in  both  length  and  direction. 

It  is  evident  that  the  direction  of  PlP2  will  not  be  changed  if 
a,  b,  and  c  are  multiplied  by  the  same  number;  in  other  words, 
the  ratios  of  the  components  are  the  essential  elements  in  fixing 
the  direction  of  the  line.  We  shall  accordingly  speak  of  a 
straight  line  as  having  the  direction  a:  b  :  c. 

On  the  other  hand,  the  length  of  the  line  does  depend  upon 
the  values  of  a,  b,  and  c,  for 


PXP  =  Vy^/+i^r+^s'2  =  Va2  +  b2  +  c\  (1 ) 

147.  Distance  between  two  points.  An  important  application 
of  (1),  §110,  is  in  finding  the  distance  between  two  points 
1\{.<\,  /jx,  zj  and  /._!(•'".,'  //.,,  ?.,).  Referring  to  fig.  198,  we  have, 
by   §U5,  a  =  P1Q=x2-zl1  b=P1R  =  y%-yv   c=P1S  =  za-z1; 

whence         ^  =  V(.r,  -  .r^ +(^- yj2 +(*,- zl}2.  (1) 

Ex.  1.  Find  the  Length  of  the  straight  line  joining  the  points  (1,  -_\ 
-1)  and  (8,  —  1,  3). 

The  required  length  is 


V(:S  -  I)-  +  (-  1  -  2)*  +  (3  +  1)-  =\  29. 

Ex.  2.  Find  a  point  Vll  units  distant  from  each  of  the  three  points 
(1,0,3),  (2,  -1,  1),  (3,  1,  •_'). 

Let  P(x,  y,  c)  be  the  required  point. 
Then  (x  -  l)2  +  (//  -  0)a  +  (z  -  3)2  =  14, 

(x-2)*  +  (y  +  l)*  +  (z-iy  =  U, 
(*-8)»  +  (y-l)»  +  (*-2)»  =  14. 
Solving  these  three  equations,  we  determine  the  two  points  (0,  2,  0)  and 
(4,  -  2,  4). 

Ex.  3.    Find  the  equation  of  a  sphere  of  radius  r  with  its  center  at 

Pi(*vyi.*i). 

If  P  (x,  y,  z)  is  any  point  of  the  sphere, 

(z_Xi)2  +  (y_yi)a  +  (z_Zi)a  =  r3. 

Conversely,  if  P (x,  y,  z)  is  any  point  the  coordinates  of  which  satisfy 
this  equation,  P  is  at  the  distance  r  from  Pv  and  hence  is  a  point  of  the 
sphere.    Therefore  this  is  the  required  equation  of  the  sphere. 


314 


SPACE  GEOMETRY 


148.  Direction  cosines.  If  we  denote  by  a,  /3,  and  7  the  angles 
which  a  straight  line  makes  with  the  positive  directions  of  the 
coordinate  axes  OX,  OY,  and  OZ  respectively,  the  cosines  of 
these  angles,  i.e.  cos  a,  cos  ft,  cosy  are  called  the  direction  cosines 
of  the  line. 

If  the  line  is  drawn  through  the  origin,  as  in  fig.  199,  it  is 
evident  that  the  same  straight  line  makes  the  angles  a,  /3,  7  or 
17  —  a,  ir  —  /3,  17  —  7  with  the  coordinate 
axes,  according  to  the  direction  in  which 
the  line  is  drawn.  Hence  its  direction 
cosines  are  either  cos  a,  cos/3,  cos  7  or 
—  cosa;,  —  cos/3,  —cos  7.  Hence  the 
straight  line  can  have  but  one  set  of 
direction  cosines  after  its  direction  has 
been  chosen. 

The   direction    cosines   may  be    deter- 
mined directly  from  the  components  of  the  line ;  for,  referring 
to  fig.  198,  we  see  that 


PQ 


or 


cos  7  = 


a 

Va2  +  &2  +  c2' 

c 
Va2  +  62  +  6"2' 


1]E 

cos/3  = 


PS 


-^-1     COS/3  =  — ,     C0S7  =  — — ,  CI) 

P,P2  P,P2  7        JJJJ  W 

b 


Va2  +  b%  +  c* 


(2) 


Squaring  and  adding  equations  (2),  we  have 

cos2a-f  cos2/3  +  cos27=l;  (3) 

that  is,  the  sum  of    the  squares  of  the  direction  cosines  of  any 
straight  line  is  always  equal  to  unity. 

It  follows   that    the   direction   cosines   of   any  line    are  not 
independent  quantities. 

Ex.    Since  the  length  of  the  line  of  Ex.  1,  §  147,  is  V29,  and  its  respec- 
tive components  are  2,  —  3,  and  4,  it  follows  that  its  direction  cosines  are 

2  3  4 

V29'       V29'  V29 


ANGLE  315 

149.  Angle  between  two  straight  lines.  Given  two  straight 
lines  having  the  respective  directions  ax :  bx :  cx  and  a% :  b2 :  cn. 
If  they  are  drawn  from  a  common  point  P  (x,  y,  z)  (fig.  200), 
let  the  segment  of  the  first  line  extend  to  Px  and  the  segment  of 
the  second  line  extend  to  P2,  so  that  the  coordinates  of  Px  are 
x  4-  dji  y  +  bx,  and  z  +  et,  and  the  coordinates 
of  P2  are  x  +  a2,  y  +  ?>„,  and  z  +  <-n.  It  follows 
that  the  components  of  PXP2  are  an—  a±,  b2  —  b^ 
and  cn  —  Cj. 

Then  if  6  is  the  angle  between  these  two 
lines,  we  have,  by  trigonometry, 

■         PP2  +  PP?-T\p;      _ 

cos  6  =  — *- i-2-.    rn 

2PP,.PP2  K  } 

But  Pli2=a;  +  b:  +  e:,  P' 

_,  Fig.  200 

Pi^O* +  &»  +  <& 

whence,  by  substitution  in  (1)  and  simplification, 

P.nsfl=  «,«,+  ¥,+  *!',  (2) 

V«f  +  ft»+«fVo£  +  tf  +  <£ 

If  cosax,  eos/3^  COS7J  are  the  direction  cosines  of  /'/;.  and  cos«2, 
cos/30,  C0S72  are  the  direction  cosines  of  J'J.U  formula  (2)  may 
be  written,  by  (2),  §  148, 

cos  6  =  cos  «1  cos  a2  +  cos  /Sx  cos  /3o  +  <•< is  7t  c< H3  7.,.         (3) 

If  the  lines  are  perpendicular  to  each  other,  cos  0  —  0,  and 
(2)  and  (3)  reduce  respectively  to 

aia2+I,A+,\('-Z=0  (4) 

and  cos  ai  cos  «2  +  cos  fix  cos  /?2  +  cos  71  cos  7,  =  0.  (5) 

If  the  lines  are  parallel  to  each  other, 

cos  ax  =  cos  «„,     cos  /3X  =  cos  /32,     cos  7X  =  cos  7„, 
whence  it  is  easily  shown  that 

?i  =  h  =  ?i.  (6) 

a.,      b.,      <?., 


316  SPACE  GEOMETRY 

150.  Direction  of  the  normal  to  a  plane.  LetiJO&n  yv  ^)  and 
%(xa,  i/.,,  z.2)  be  any  two  points  of  the  plane 

Ax  +  By  +  Cz  +  I)=Q.  (1) 
Substituting  their  coordinates  in  (1),  we  have 

Axx  +  Byx  +Cz1  +  D  =  Q  (2) 

and  Ax2  +  By„  +  Cz„  +  D  =  0.  (3) 

Subtracting  (2)  from  (3),  we  have 

A  jg  -  g|  +  2?  <£  -  yx)  +  C  (z,  -  dj  =  0,  (4) 

whence,  by  (4),  §  149,  the  direction  A:  B:C  is  normal  to  the 
direction  x2  —  xx :  «/2  —  yx :  z2  —  zx.  But  the  latter  direction  is  the 
direction  of  any  straight  line  of  the  plane.  Hence  the  direc- 
tion A  :  B :  C  is  the  direction  of  the  normal  to  the  'plane  Ax  4-  By 
+  Cz  +  D  =  0. 

151.  Equation  of  a  plane  through  a  given  point  perpendicular 
to  a  given  direction.  Let  the  plane  pass  through  a  given  point 
^i(-ri'  V-a  ~i)  perpendicular  to  a  straight  line  having  a  given 
direction  A:B:  C.  Let  P(x,  y,yz}  be  any  point  of  the  plane. 
Then  x  —  x1:  y  —  yl:  z  —  zx  is  ^he  direction  of  iJP,  i.e.  is  the 
direction  of  any  straight  line  through  Px  in  the  plane. 

Since  a  perpendicular  to  a  plane  is  perpendicular  to  every 
line  in  the  plane,  it  follows  that 

^ A  (x  -X^  +  B(y-  yj)  +  C(z  -  zj  =  0, 
which  is,  accordingly,  the  required  equation  of  the  plane. 

Since  every  plane  may  be  determined  in  this  way,  and  this 
equation  is  a  linear  equation,  it  follows  that  every  plane  may 
be  represented  by  a  linear  equation. 

Ex.    Find  the  equation  of  a  plane  passing  through  the  point  (1,  2,  1) 
and  normal  to  the  straight  line  having  the  direction  2  : 3  :  —  1. 
The  equation  is 

2(x-l)  +  3(y-2)-l(2-l)  =  0, 

or  2x  +  3y-z-7  =  0. 

152.  Angle  between  two  planes.    Let  the  two  planes  be 

Af  +  Bjf+Cp  +  D^O,  (1) 

Ajc  +  B„y  +  C2z  +  Z>2  =  0.  (2) 


STRAIGHT  LINE  317 

The  angle  between  these  planes  is  the  same  as  the  angle 
between  their  respective  normals,  the  directions  of  which  are 
respectively  the  directions  Ax:  B1:  C\  and  A.,:  />'.,:  Cg.  Hence,  if 
6  is  the  angle  between  the  two  planes,  by  (2),  §  149, 

,  AxA«  +  B,B„+C,Cn 

cos  6  =  1    -        1    ;        1  =  • 

Va?  +  a;  +  c*  V  a*  +  J5aa  +  cy2 

The  conditions  for  perpendicularity  and  parallelism  of  the 
planes  are  respectively 

i  A      »,      Ci 

153.  Equations  of  a  straight  line.  In  space  of  three  dimen- 
sions a  single  equation  in  general  represents  a  surface;  hence,  in 
general,  a  curve  cannot  he  represented  by  a  single  equation.  A  curve 
may,  however,  be  regarded  as  the  line  of  intersection  of  two 
surfaces.  Then  the  coordinates  of  every  point  of  the  curve 
satisfy  the  equations  of  the  surfaces  simultaneously  :  and,  con- 
versely, any  point  the  coordinates  of  which  satisfy  the  equations 
of  the  surfaces  simultaneously  is  in  their  curve  of  intersection. 
Hence,  in  general,  the  locus  of  two  simultaneous  equations  in  x,  //, 
and  z  is  a  curve. 

In  particular,  the  locus  of  the  two  simultaneous  linear  equations 

\x+Bxy  +  C'12+/>1=0, 
A2x  +  B„y  +  C^+J)2=0, 

is  a  straight  line,  since  it  is  the  line  of  intersection  of  the  two 
planes  respectively  represented  by  the  two  equations. 

We  will  now  find  the  equations  of  the  straight  line  determined 
by  two  points,  and  the  equations  of  the  straight  line  passing  through 
a  known  point  in  a  given  direction. 

154.  Straight  line  determined  by  two  points.  Let  the  given 
points  be  B,(xv  yv  zx)  and  B,(x2,  y.2,  z2).  Then  the  direction  of 
1{B  is  #2  —  .r, :  //.,  —  yl :  z.,—  zv  Let  P(x,  y,  z)  be  any  point  of  the 
line.    Then  the  direction  of  P^P  is  x  —  x\  :  y  —  yx :  z  —  zx. 


318 


SPACE  GEOMETRY 


Since  I[P  and  I(P1  are  parts  of  the  same  straight  line,  and 
hence  parallel,  it  follows  that 

x  ~  x\  _  V  ~  V\  _  z  ~  z\ 

Here  are  but  two  independent  equations  in  x,  y,  and  z.  This 
result  proves  the  converse  of  the  statement  above,  that  two 
linear  equations  always  represent  a  straight  line ;  for  we  have 
any  straight  line  represented  by  two  linear  equations. 

It  is  to  be  noted  that,  if  in  the  formation  of  these  fractions 
any  denominator  is  zero,  the  corresponding  component  is  zero, 
and  the  line  is  perpendicular  to  the  corresponding  axis. 

Ex.  Find  the  equations  of  the  straight  line  determined  by  the  points 
(1,  5,  -  1)  and  (2,  -  3,  -  1). 

x  —  1  _    y  —  5     _    z  +  1 
2^T~ -3- 


1  +  1 


Hence  the  two  equations  of  the  line  are  z  + 1  =  0,  since  the  line  is  par- 
allel to  the  XOY  plane  and  passes  through  a  point  for  which  z  =  —  1,  and 
8  x  +  y  — 13  =  0,  formed  by  equating  the  first  two  fractions. 

155.  Straight  line  passing  through  a  known  point  in  a  given 
direction.  If  the  direction  of  the  line  is  given  as  a :  b  :  c,  the 
equations  of  the  line  are  evidently 

x  —  x,      y  —  y,      z  —  z.  ^ . 

(!) 


a  b  c 

for  in  the  formula  of  §  154  we  may  place 

If  the  direction  of  the  line  is  given  hi        / 
terms  of  its  direction  cosines,  the  derivation     y        Fig.  201 
of  the  equations  is  as  follows : 

Let  P1(xv  yv  2j)  (fig.  201)  be  a  known  point  of  the  line,  and 
let  I,  m,  and  n  be  its  direction  cosines.  Let  P(x,  y,  z)  be  any 
point  of  the  line.  On  P^P  as  a  diagonal  construct  a  parallele- 
piped as  in  §  146.    Then  if  we  denote  P^P  by  r,  we  have 

PxQ=lr,     %B=  mr,     %S=  nr. 
But  PxQ=x—xvl     PR—y  —  yv     I£S=z  —  e1, 

whence  x—  x,=  lr.     y  —  y=  mr,     z  —  2  =  nr. 


STRAIGHT  LINE  319 

Eliminating  r  from  these  last  three  equations,  we  have 

x  -  *i  _  y  -  Vx  _ z  -  zi ,  ^2) 


l  m  n 

which  are  but  two  independent  linear  equations. 

156.  Determination  of  the  direction  cosines  of  a  straight  line. 
If  the  equations  of  the  straight  line  are  hi  any  one  of  the  forms 
of  §§  154  and  155,  the  determination  of  the  direction  cosines  is 
very  easy,  for  the  denominators  of  the  fractions  in  those  formu- 
las are  either  the  direction  cosines  of  the  line  or  else  give  the 
components  for  the  line,  from  which  the  direction  cosines  are 
quickly  computed. 

If  the  equations  of  the  straight  line,  however,  are  in  any 

other  form,  as  „      .    _        ^       . 

./,./• +  /;i//  +  r,  +  /,i  =  o,  (1) 

J,r  +  n  ji  +  cj  4  Da  =  0,  (2) 

let  its  direction  cosines  be  /,  m,  and  n.  Since*  the  line  lies  in 
both  planes  (1)  and  (2),  it  is  perpendicular  to  the  normal  to 
each.    Therefore,  by  (4),  §  149, 

A  J  4  Bxm  4-  Cxn  =  0, 

AJ,  4  Btm  4  Can  =  0  ; 

also  l*  +  m*  +  n*=l.  (§148) 

Here  are  three  equations  from  which  the  values  of  I,  m,  and  n 
may  be  found. 

Ex.  Find  the  direction  cosines  of  the  straight  line  2z+3y+«—  4  =  0, 
4  x  +  >/-  z  +  7=0. 

The  three  equations  for  I,  m,  and  n  are 

2  Z  +  3  m  4  n  =  0, 
4  I  +  m  —  n  =  0, 

/-  +  m*  +  n"  =  1, 
the  solutions  of  which  are 

z  =  _2_; 3_?  5 

V38'                    V38'  V3s' 

,_        2                        3  5 

V38                   V38  V38 


320 


SPACE  GEOMETRY 


Since  cos (180°  —  <f>)  =  —  COS <ji,  it  is  evident  that  if  the  angles  corre- 
sponding  to  the  first  solution  are  a,,  j3v  yv  the  angles  corresponding  to 
the  second  solution  are  180°  —  ap  180°  —  fiv  180°  —  yr  Since  these  two 
directions  are  each  the  negative  of  the  other,  it  is  sufficient  to  take  either 
solution  and  ignore  the  other. 

157.  Distance  of  a  point  from  a  plane.  Let  it  be  required 
to  find  the  perpendicular  distance  from  the  point  -^(a^,  yx,  aQ 
to  the  plane 

Ax+By  +  Cz+P>  =  0.  (1) 

From  P.  (fig.  202)  draw  the 
required  perpendicular  PXN 
and  also  a  line  parallel  to 
the  axis  of  z  and  let  it  cut 
the  plane  in  R.  Then  for  the 
point  R,  x  =  Xf  y  =  yx,  and  z 
is  determined  from  the  equa- 
tion of  the  plane  as 

-Ax1-By1-D 


Hence 


RP1 


_Ax1+By1  +  Cz1+I)_ 
C 
But  P1N=RP1  cos  7  where  <y  is  the  angle  R^N  which  is  equal 
to  the  angle  made  by  the  normal  to  the  plane  with  the  line  OZ. 
Then,  by  §  150,  c 


cos  7 


Hence 


PXN  =  ± 


VA*  +  B1  +  Cz 
Ai\+By,  +  Cz,+D 


^A'+B^  +  C* 

is  the  magnitude  of  the  required  distance,  being  positive  for  all 
points  on  one  side  of  the  plane  and  negative  for  all  points  on 
the  other  side.  If  we  choose,  we  may  take  the  sign  of  the 
radical  always  positive,  in  which  case  we  can  determine  for 
which  side  of  the  plane  the  above  result  is  positive  by  testing 
for  some  one  point,  preferably  the  origin. 


STRAIGHT  LINE  321 

Ex.  Find  the  distance  of  the  point  (1,  2,  1)  from  the  plane  2  x  —  3  y 
+  6  s  +  11  =  0.    The  required  distant ■  is 

2(l)-3(2)+0(l)  +  U  = 

7 

Furthermore  the  point  is  on  the  same  side  of  the  plane  as  the  origin,  for 
if  (03  0,  0)  had  been  substituted,  the  residt  would  have  been  2,  i.e.  of  the 
same  sign  as  2f . 

158.  Problems  on  the  plane  and  the  straight  line.  In  this 
article  we  shall  solve  some  problems  illustrating  the  use  of 
the  equations  of  the  plane  and  the  straight  line. 

1.  Plane  through  a  given  line  and  subject  to  one  other  condition. 
Let  the  given  line  be 

As  +  Bj-tCf  +  D^O,  (1) 

Ap  +  B#  +  Cf+D=Q.  (2) 

Multiplying  the  left-hand  members  of  (1)  and  (2)  by  l\  and  Jca 
respectively,  where  kx  and  /-.,  are  any  two  quantities  independent 
of  x,  y,  and  2,  and  placing  the  sum  of  these  products  equal  to 
zero,  we  have  the  equation 

\(A ,/•  +  //,//  +  Cxz +2>x)  +  fca0 Kx  +  BJf  +''/+  ",)  =  o.   (3) 

Equation  (3)  is  the  equation  of  a  plane,  since  it  is  a  linear  equa- 
tion, and  furthermore  it  passes  through  the  given  straight  line, 
since  the  coordinates  of  every  point  of  that  line  satisfy  (3)  by 
virtue  of  (1)  and  (2).  Hence  (3)  is  the  required  plane,  and 
it  may  be  made  to  satisfy  another  condition  by  determining  the 
values  of  k^  and  /-.,  appropriately. 

Ex.  1.  Find  the  equation  of  the  plane  determined  by  the  point  (0,  1.  0  | 
and  the  line  4  x  +  3  y  +  2z  -  4  =  0,  2x  — 11  y  -  I  :  -  12  =  0. 

The  equation  of  the  required  plane  may  be  written 

/M  (4x  +  3  y  +  2  z  -  4)  +  /,-,( 2  x  - 11  y  -4a-  12)  =  0.  (1 ) 

Since  (0,  1,  0)  is  a  point  of  this  plane,  its  coordinates  satisfy  (1),  and 
hence  fcj  +  23  / ,  =  0,  or  l\  =  -  23  I:,. 

Substituting  this  value  of  /c'x  in  (1),  and  reducing,  we  have  as  the  required 
e(luatiuu>  9x  +  8y  +  5z-8  =  0. 


322  SPACE  GEOMETRY 

Ex.  2.  Find  the  equation  of  the  plane  passing  through  the  line  4  x  +  3  y 
+  2  z  —  4  =  0,  2  .r  —  11  y  —  4  s  — 12  =  0,  and  perpendicular  to  the  plane 
2  x  +  y-2z  +  l  =  0. 

The  equation  of  the  required  plane  may  be  written 

^(4  x  +  3y  +  2z-i)  +  k2(2  x  -  11  y  -  4  z  -  12)  =  0,  (1) 

or  (4  kx  +  2  fc2) x-  +  (3  ^  - 11  £a) y  +  (2  hx  -  4  k2)z  +  (-  4 ^  -  12  fra)  =  0. 

Since  this  plane  is  to  be  perpendicular  to  the  plane  2  x  +  y  —  2z  +  l  =  0, 

2  (4  fcj  +  2  fc2)  +  1  (3  k,  -  11  ifc2)  -  2  (2  ^  -  4  k2)  =  0, 

whence  k„=  —  7  kv 

Substituting  this  value  of  k2  in  (1),  and  reducing,  we  have  as  the  required 

equation, 

x  —  8y  —  32  —  8  =  0. 

2.  Plane  determined  by  three  points.  If  the  equations  of  the 
straight  line  determined  by  two  of  the  points  are  derived,  we 
may  then  pass  a  plane  through  that  line  and  the  third  point,  as 
in  Ex.  1.    The  result  is  evidently  the  required  plane. 

Ex.  3.  Find  the  equation  of  the  plane  determined  by  the  three  points 
(1,  1,  1),  (-1,  1,  2),  and  (2,  -  3,  -1). 

The  equations  of  the  straight  line  determined  by  the  first  two  points  are 

x - 1         y - 1       z-1 


-1-1      1-1      2-1 

which  reduce  toy  —  1  =  0,       x  +  2z  —  3  =  0. 

The  equation  of  the  required  plane  is  now  written  in  the  form 
k1(y-l)  +  k2(x  +  2z- 3)^=0. 

Substituting  (2,  —  3,-1)  in  this  equation,  we  have 
-  4  kt  -  3  fcg  =  0,    or     &2  =  —  |  kv 

Substituting  this  value  of  k2  in  the  equation  of  the  plane,  and  simplify- 
ing, we  have  as  our  required  equation, 

4a:-3#  +  8z-9  =  0. 

159.   Space  curves.     We  saw  in  §  153  that,  in  general,  the 
locus  of  two  simultaneous  equations  in  x,  y,  and  z  is  a  curve  — 


CURVES  323 

the  curve  of  intersection  of  the  surfaces  represented  by  the 
equations  taken  independently. 

Let  /1(a;,y,2)=0,    /,(*yf«)=0,  (1) 

be  the  two  equations  of  a  space  curve. 

If  we  assign  a  value  to  one  of  the  coordinates  in  equations 
(1),  as  x  for  example,  there  are  two  equations  from  which  to 
determine  the  corresponding  values  of  y  and  z,  in  general  a 
determinate  problem.  But  if  values  are  assigned  to  two  of  the 
coordinates,  as  x  and  y,  there  are  two  equations  from  which  to 
determine  a  single  unknown,  z,  a  problem  generally  impossible. 
Hence  there  is  only  one  independent  variable  in  the  equations 
of  a  curve. 

In  general,  we  may  make  x  the  independent  variable  and 
place  the  equations  in  the  form 

y  =  cj>1(x~),     z  =  4>2(x),  (2) 

by  solving  the  original  equations  (1)  of  the  curve  for  y  and  z 
in  terms  of  x.  The  new  surfaces,  y  =  <f>1(z),  z  =  (f>j.r),  deter- 
mining the  curve,  are  cylinders  ($  142),  with  elements  parallel 
to  OZ  and  0  Y  respectively.  The  equation  y  =  <\>x(x)  interpreted 
in  the  plane  XOY  is  the  equation  of  the  projection  (§145)  of 
the  curve  on  that  plane.  Similarly,  the  equation  z  =  <f>i(x)i 
interpreted  in  the  plane  ZOX,  is  the  equation  of  the  projection 
of  the  curve  on  that  plane. 

Hence,  to  find  the  projection  of  the  curve  (1)  on  the  XOY  plane 
we  eliminate  z  from  the  two  equations. 

Similarly,  to  find  the  projection  on  the  XOZ  plane  we  eliminate 
y,  and  to  find  the  projection  on  the  YOZ  plane  we  eliminate  x. 

Finally,  the  three  equations 

*=/x(9.   y-/.(0.   a=/»(0  (3) 

are  parametric  equations  of  a  curve.  They  may  generally  be 
put  in  the  form  y  =  <f>1(_x),  3  =  <£.,(#),  by  eliminating  t  from 
the  first  and  second  equations,  and  from  the  first  and  third 
equations. 


324 


SPACE  G  EOMETK  Y 


Ex.  The  space  curve  called  the  helix  is  the  path  of  a  point  which  mows 
around  the  surface  of  a  right  circular  cylinder  with  a  constant  angular 
velocity  and  at  the  same  time  moves  parallel  to  the  axis  of  the  cylinder 
with  a  constant  linear  velocity. 

Let  the  radius  of  the  cylinder  (fig.  203)  be  a,  and  let  its  axis  coincide 
with  OZ.  Let  the  constant  angular  velocity  be  w  and  the  constant 
linear  velocity  be  v.  Then  if  6  denotes  the  angle  through  which  the 
plane  ZOP  has  swung  from  its  initial  position  ZOX,  the  coordinates 
of  any  point  P  (x,  y,  z)  of  the  helix  are  given  by  the  equations 


I>nt  6  =  oit,  and  accordingly  we  may  have  as  the 
parametric  equations  of  the  helix, 

x  =  a  cos  (at, 
y  =  a  sin  wt, 

t  being  the  variable  parameter. 
n 
Or,    since    t  =  —■>    we    may   regard    6    as    the 

CO 

variable  parameter,  and  the  equations  are 
x  =  a  cos  6, 
y  =  a  sin  6, 
z  =  led, 

where  k  is  the  constant  —  • 


Fig.  203 


160.  Direction  of  space  curve  and  element  of  arc.  Let  P(.r,  y,  z) 
be  any  point  of  a  curve,  and  Q  (x  +  Ax,  y  +  Ay,  z  +  Az)  be  any 
second  point  of  the  curve.  Then  the  direction  cosines  of  the 
chord  PQ  are 


A.r 


Ay 


Az 


'  Vaz2  +  Ay2  +  A/     y/Ax2  +  Ay2  +  Az'2     y/Ax*  +  Ay2  +  Az'2 

As  the  point  Q  approaches  the  point  P  along  the  curve, 
Ax,  Ay,  and  Az  each  approach  zero  as  a  limit,  and  the  direction 
cosines  of  the  chord  PQ  approach  the  direction  cosines  of  the 
tangent  to  the  curve  at  P  as  limits.  To  determine  these  limits, 
denote  by  s  the  distance  of  the  point  P  from  some  fixed  point 


CURVES  325 

of  the   curve,  s  being  measured   along  the   curve.     Then   the 
arc  PQ  =  As;   and  As  =  0  as  PQ  approaches  the  tangent. 

Ax  Ax  As 


Now 


Vax2+a/+az2    As  VZ7+a/Ta72' 

Ax  dx 


whence  Lim 

Asi0VAl-2+Ay  +  A22      ds 

for  Lim  — ^==^===^  =  1. 

WAx'+A/  +  Az2 

Proceeding  in  the  same  way  with  the  other  two  ratios,  we 

,         dx    dy    dz        ,,      ,.      ,.  .  .  , . 

nave  — ,  -f-,  —  as  the  direction  cosines  of  the  curve  at  any 
as     as     as  J 

point,  since  the  directions  of  the  tangent  and  the  curve  at  any 

point  are  the  same. 

whence  ds  =  Vdx'2  +  dy1  +  dz'2,  (1 ) 

a  formula  for  the  differential  of  the  arc  of  any  space  curve. 

It  also  follows  from  (1)  that  we  may  speak  of  the  direction 
of  the  curve  as  the  direction  dx :  dy  :  dz. 

Ex.  1.    Find  the  direction  of  the  helix 

x  =  (i  cos  6,     y  =  a  sin  6,     z  =  kd, 
at  the  point  for  which  0  =  <>. 

Here  dx  =  —  a  sin  0  dO,  dy  —  a  cos  6  dO,  d:  =  k  d$.  Therefore,  at  the  point 
for  which  $  =  0,  the  direction  is  the  direction  0  :  a  d6  :  k  dd,  and  the  direction 

a  k 

cosines  are  0, 


Va-  +  k-   Vos  +  k* 

Ex.  2.    Find  the  length  of  an  arc  of  the  helix  corresponding  to  an 
increase  of  2  ir  in  6. 

Using  the  values  of  dx,  dy,  and  dz  found  in  Ex.  1,  we  have 


ds  =  Va2  +  P  dd ; 


whence  s  =   /  Va2  +  k2d$ 


e,  +  2tt 


2  7rV«2  +  k-. 


326  SPACE  GEOMETRY 

161.  Tangent  line  and  normal  plane.  If  J^(xv  yv  sx)  is  the  point 
of  tangeney,  the  equations  of  the  tangent  line  are,  by  (1),  §  155, 

x  —  x,      y  —  il      z  —  2, 

*  =  2 — -n  = 1,  f-jN 

dxx  dyx  dzl  ^  J 

where  dxx,  dyv  dzx  are  the  respective  values  of  dx,  dy,  dz  at  the 
point  Pv 

The  plane  perpendicular  to  a  tangent  line  at  the  point  of 
tangeney  is  called  the  normal  plane  to  the  curve. 

By  §  151,  the  equation  of  the  normal  plane  to  the  curve  at  ij*  is 

dxx(x  -  aQ  +  dy^y  -  yj  +  dz^z  -  sQ  =  0.  (2) 

Ex.  Find  the  equations  of  the  tangent  line  and  the  equation  of  the 
normal  plane  to  the  helix 

x  =  a  cos  9,     y  =  a  sin  6,     z=  kd 

at  the  point  for  which  6  =  0.  Here  xx  —  a,  yx  =  0,  zx  =  0,  and  dx1  =  0, 
dij  =  add,  d~1  =  kdd.    Hence  the  equations  of  the  tangent  line  are 


x  —  a      y  —  0       z  —  0 
0      ~   ud8    ~  kdd  ' 

which  reduce  to 

x  —  a  =  0,     ky  —  az  =  0 

The  equation  of  the  normal  plane  is 

0(x  -  a)  +  ad$(y  -  0)  +  kdd(z  -  0)  =  0, 
which  reduces  to  ay  +  kz  =  0. 

PROBLEMS 

1.  Describe  the  surface  if  —  4//  —  2x  —  0. 

2.  Describe  the  surface  y(z  —  1)  =  1. 

3.  Write  down  the  equation  of  a  right  circular  cylinder  of  radius  a. 

4.  Show  that  the  surface  ax  +  by  =  cs2  is  a  cylinder,  and  describe 
its  directrix  and  generatrix. 

5.  Describe  the  surface  x2  +  y2  +  z2  —  6  z  =  0. 

6.  Describe  the  surface  xyz  =  a3. 

7.  Describe  the  surface  9x2  +  4  «2  =  12  #  -  24. 

8.  Describe  the  locus  of  the  equation  x  —  (z  +  2)2  =  0. 


PROBLEMS  327 

9.  Show  that   the    surface   (ax  4-  byf  =  cz   is   a  cylinder,  and 
describe  its  directrix  and  generatrix. 

10.  Describe  the  surface  36  x2  —12x  +  <dif  +  ±z1  =  0. 

11.  All  sections  of  a  given  right  cylinder  made  by  planes  parallel 
to  the  plane  XOZ  are  ellipses  of  which  the  longest  chord  is  10  in. 
and  the  shortest  is  8  in.    What  is  the  equation  of  the  cylinder  '.' 

12.  Describe  the  surface  x*  4-  if  +  z3  =  '/f. 

13.  Show  that  the  surface  z  =  a  —  vVJ  4-  //-  is  a  cone  of  revolul  ion, 
and  find  its  vertex  and  axis. 

14.  Find  the  equation  of  a  prolate  spheroid,  i.e.  the  surface  gen- 
erated by  revolving  an  ellipse  about  its  major  axis. 

15.  Find  the  equation  of  an  oblate  spheroid,  i.e.  the  surface  gen- 
erated by  revolving  an  ellipse  about  its  minor  axis. 

16.  Describe  the  locus  of  the  equation  xa  4-  -1  xy  4-  4  if  —  4 .-.-  =  0. 

8  a9 

17.  Describe  the  surface  z  =  —, 5 : — ;• 

■'-  +  //-  4-  4  it- 
is.   Find  the  equation  of  the  cone  of  revolution  formed  by  revolving 
the  line  z  =  2x  about  OX  us  an  axis. 

19.  Describe  the  locus  of  the  equation  -.>■'-  —  3x  —  2  =  0. 

20.  Find  the  equation  of  a  parabolic  cylinder  the  elements  of  which 

are  parallel  to  OX  and  the  directrix  of  which  is  in  the  plane  YOZ. 

//-        -A 

21.  Describe  the  surface  '—  +  ■—  +  —.=1. 

n-  Ir  r4 

22.  Describe  the  surface  if  —(2a  —  //)(.-.'-  +  x2)—  0. 

23.  Describe  the  surface  x'2  —  f  —  2  ./■  +  4  y  —  0. 

24.  Find  the  equation  of  the  cone  of  revolution  formed  by  revolv- 
ing the  line  3//  =  2x  +1  about  the  line  y  =1  in  the  plane  A'OT  as 
an  axis.    What  are  the  coordinates  of  the  vertex  of  the  cone  ? 

25.  Show  that  the  surface  ./•'■  +  2  f  -  3  z2  +  2  x  - 12 y  +  1 2  z  +  7  =  0 
is  a  cone  with  its  vertex  at  the  point  (—1,  3,  2).  What  are  its  cross 
sections  made  by  plunes  parallel  to  the  plane  XOY? 

26.  Describe  the  surface  (x  —  a)x2  +  (•'•  +  «)  if  +  -~")  =  0. 

27 .  Find  the  equation  of  the  ring  surface  formed  by  revolving  the 
ellipse  4-^=1  (a  >  b)  about  <>Y  as  an  axis. 


328  SPACE  GEOMETRY 

28.  Describe  the  surface  4  s2  =  y2(9  —  a;2). 

29.  If  P(x,  y,  z)  is  situated  on  the  straight  line  drawn  from 
P&v  I/v  -x)  to  I\(x2,  yi}  z2)  so  that  P1P  =  k(PxP£,  prove  that 

x  =  x1  +  k(x2-x1),     y  =  y1  +  A;(y3-y]),     z  =  z1  +  k  («a  -  «x). 

i    //i  +  &   *i  +  *a\  ig  the  middle  point  of 


2  2  2 

the  straight  line  joining  P1(a;1,  y^  zj  and  P2(x2,  y2,  z2). 

31.  Find  the  equation  of  the  sphere  constructed  on  the  straight 
line  joining  (3,  —  1,  3)  and  (5,  3,  5)  as  a  diameter. 

32.  Find  a  point  of  the  plane  x+3y+z=0  equally  distant 
from  the  three  points  (1,  1,  1),  (0,12,  1),  (2,  1,  2). 

33.  Find  the  points  distant  5  from  the  points  (—2,  —  2,  1), 
(3,  -  2,  6),  (3,  3,  1). 

34.  Find  the  point  of  the  plane  x  +  2y  +  3p  —  6  =  0  equally 
distant  from  the  points  where  the  plane  is  pierced  by  the  three 
coordinate  axes. 

35.  Find  the  equation  of  the  sphere  passing  through  the  points 
(-  1,  1,  -  5),  (-  2,  4,  3),  (-  5,  0,  -  2),  (7,  1,  -  1). 

36.  A  point  moves  so  that  its  distances  from  two  fixed  points  are 
in  the  ratio  k.  Prove  that  its  locus  is  a  sphere  or  a  plane  according 
as  k  =£  1  or  k  =  1. 

37.  Prove  that  the  locus  of  points  from  which  tangents  of  equal 
length  can  be  drawn  to  two  given  spheres  is  a  plane  perpendicular 
to  their  line  of  centers. 

38.  A  straight  line  makes  the  same  angle  with  the  three  coordi- 
nate axes.   What  is  that  angle  ? 

39.  Prove  that  a  straight  line  can  make  angles  60°,  45°,  60° 
respectively  with  the  coordinate  axes. 

40.  Find  the  direction  cosines  of  the  straight  line  determined  by 
the  points  (1,  3,  5),  (2,  -  1,  4). 

41.  A  straight  line  makes  an  angle  of  30°  with  OX  and  equal 
angles  with  O  Y  and  OZ.    What  is  its  direction  ? 

42.  Find  the  angle  between  the  two  straight  lines  joining  the 
origin  to  the  points  (1,  2,  1)  and  (3,  —  1,  3). 


PROBLEMS  329 

43.  Prove  that  the  three  points  (5,  3,  -  2),  (4,  1,  - 1),  (2,  -  3,  1) 
lie  on  one  straight  line. 

44.  Through  the  point  (1,  —  3,  1)  of  the  straight  line  having  the 
direction  1:2:3a  straight  line  is  drawn  to  the  point  (4,  2,  0).  Find 
the  angle  between  the  two  lines. 

45.  Prove  that  P^-l,  2,  1),  P2(2,  3,  5),  and  P,(4,  5,  3)  are  the 

vertices  of  a  right  triangle. 

46.  Find  the  equation  of  a  plane  passing  through  the  point 
(—  2,  3,  —  4)  parallel  to  the  plane  x  —  3y  -{-  7  z  —  11  =  0. 

47.  Find  the  equation  of  a  plane  passing  through  the  point 
(5,  —  2,  7)  equally  inclined  to  the  three  coordinate  axes. 

48.  Find  the  equation  of  a  plane  perpendicular  to  the  straight 
line  joining  the  points  (1,  3,  5)  and  (4,  3,  2)  at  its  middle  point. 

49.  Find  the  equation  of  a  plane  passing  through  the  point 
(1,  1,  2)  perpendicular  to  the  straight  line  determined  by  the  points 
(1,  -  1,  1)  and  (3,  1,  3). 

50.  What  is  the  angle  between  the  planes  2x-\-  y  —  72  +  11  =  0, 
5a5-2y  +  5«-12  =  0? 

51.  Find  the  angle  between  the  planes  3x  +  2  y  —  4  =  0,  2y  -\-3z 
+  13  =  0. 

52.  Find  the  equations  of  the  straight  line  determined  by  the 
points  (6,  2,  - 1)  and  (3,  4,  -  4). 

53.  What  are  the  equations  of  the  straight  line  determined  by 
the  points  (2,  3,  5)  and  (1,  -1,5)? 

54.  Find  the  equations  of  a  straight  line  passing  through  the 
point  (0,  3,  5)  perpendicular  to  the  plane  x  +  3y  +  5z  —  9  =  0. 

55.  A  straight  line  is  drawn  through  the  point  (4,  6,  —  2)  parallel 
to  the  straight  line  drawn  from  the  origin  to  the  point  (1,  —  5,  3). 
What  are  its  equations  ? 

56.  A  straight  line  making  angles  60°,  45°,  and  60°  respectively 
with  the  axes  of  x,  y,  and  z  passes  through  the  point  (2,  —  2,  2). 
What  are  its  equations  ? 

57.  A  straight  line  passes  through  the  point  (2,  —  5,  2)  parallel 
to  OY.    What  are  its  equations  ? 

58.  Find  the  direction  cosines  of  the  line  4cc  —  3y  —  4  =  0, 
12a;  _  3s -15  =  0. 


330  SPACE  GEOMETRY 

59.  Find  the  direction  cosines  of  the  line  3x  +  y  —  1  z  —  6  =  0, 
2a;-3y  +  4»-7=0. 

60.  Find  the  equations  of  the  straight  line  passing  through 
(1,  3,  —  5)  parallel  to  the  line  y  =  3  a-  — 14,  7  x  —  2z  =  17. 

61.  Prove  that  the  three  planes  x  —  2  y  + 1  =  0,  7  //  —  z  —  4  =  0, 
7x-  22  +  6=0  are  the  lateral  faces  of  a  triangular  prism. 

62.  Find  the  angle  between  the  line  3x  —  2  y  —  4  =  0,  y  +  3z 
+  5  =  0  and  the  plane  3  x  +  y  -  2  z  +  31  =  0. 

63.  Find  the  distance  of  the  plane  2*  +  3s +11=  0  from  the 
origin. 

64.  Find  the  locus  of  points  distant  3  from  the  plane  x  +  y  +  z 
+  3  =  0. 

65.  Find  the  locus  of  points  equally  distant  from  the  planes 
x  +  2  y  +  3  z  +  4  =  0,  x  -  2  y  +  3  z  -  5  =  0. 

66.  Find  a  point  on  the  line  3x  —  2 y  — 11=0,  2x  —  y—  z  —  5  =  0 
equally  distant  from  the  points  (0,  1,  1)  and  (1,  2,  1). 

67.  Find  the  equation  of  the  plane  passing  through  the  point 
(2,  —  3,  —  2)  perpendicular  to  the  line  2 x  +  y  —  5z  —  7=0,  y  +  2 z 
-4  =  0. 

68.  Find  the  equation  of  a  plane  four  units  distant  from  the 
origin  and  perpendicular  to  the  straight  line  through  the  origin  and 

(1,  -  5,  6). 

69.  A  straight  line  is  drawn  from  the  origin  to  the  plane  2x  +  y 
+  2  z  —  5  =  0.  It  makes  equal  angles  with  the  three  coordinate  axes. 
Find  its  length. 

70.  Find  the  coordinates  of  a  point  on  the  straight  line  determined 
by  (- 1,  0,  1)  and  (1,  2,  3)  and  3  units  distant  from  (2,  - 1,  1). 

71.  Find  the  foot  of  the  perpendicular  drawn  from  (3,  —  2,  0)  to 
the  plane  2x  +  y  —  4«+17=0. 

72.  Find  the  length  of  the  projection  of  the  straight  line  joining 
the  points  (1,  2, 1)  and  (2,  —1,  2)  upon  the  straight  line  determined 
by  the  points  (2,  1,  3)  and  (4,  4,  6). 

73.  Find  the  equation  of  the  plane  determined  by  the  three  points 
(1,  3,  -  2),  (0,  2,  - 10),  and  (-  2,  4,  -  6).      ' 

74.  Find  the  direction  of  the  normal  to  the  plane  determined  by 
the  three  points  (1,  2,  3),  (-1,  -  2,  -  3),  (4,  -  2,  4). 


PROBLEMS  331 

75.  Find  the  point  of  intersection  of  the  lines 

fs  +  2y-3  =  0        1  f3y  +  5*  +  15  =  01 

\x  +  y-  2*- 9  =  0/  and   l3z-2*-15  =  0/ 

76.  Prove  that  the  lines 

\2x-2>/-z  +  3  =  0J  U.,-2.~-3  =  0j 

intersect  at  right  angles. 

77.  Prove  that  the  two  lines 

fx  +  2y-z  +  7  =  0      \  Ux-7y  +  8s  +  19  =  01 

\2x-y  +  2z  +  U  =  0}  an      U-3y  +  3s  +  4  =  0     J 
are  coincident. 

78.  Prove  that  the  two  lines 

r3a._2.y-7=01  (x  -  z  =  0  1 

\2  //  -  3  «  +  7  =  0  J  13  .r  -  4  //  +  3  c  -  8  =  0  J 

can  determine  a  plane,  and  derive  its  equation. 

79.  Prove  that  the  two  lines 

f2,+3//  +  4  =  01  f*  +  2y  +  *  +  2  =  0    1 

\2y  +  «  +  3  =  0    J   ,U      l2x-y  +  22-9  =  0/ 
cannot  determine  ;i  plane. 

80.  Prove  that  the  two  lines 

Cx  -  2  //  - 10  =  01  f  7a;  -  3  --11=0  1 

\4  y  -  z  + 17  =  0  J    aD<     17a;  +  14  y  -  «  +  43  =  Oj 
can  determine  a  plane,  and  derive  its  equation. 

81.  Find  the  equation  of  a  plane  passing  through  the  line 
./•+// +  3--  —  7  =  0,   3x  +  2y  —  a  =  0  and   perpendicular   to   the 

plane  2  x  +  //  -  2 .?  +  11  =  0. 

82.  Find  the  equation  of  a  plane  passing  through  the  points 
(—2,  3,  —2),  (2,  — 1,  2)  perpendicular  to  the  straight  line  deter- 
mined by  the  points  (0,  0,  0),  (1,  2,  1). 

83.  Find  the  equation  of  the  plane  determined  by  the  point 
(1,  5,  —  2)  and  the  straight  line  passing  through  the  point  (6,  —  2,  4) 
equally  inclined  to  the  coordinate  axes. 

84.  Find  the  equation  of  the  plane  passing  through  the  points 
(0,  3,  2),  (2,  -  3,  4)  perpendicular  to  the  plane  Gx  +  3y-2z  +  3  =  Q. 


332  SPACE  GEOMETRY 

85.  Find  the  equation  of  a  plane  determined  by  the  point 
(2,  3,  2)  and  the  straight  line  passing  through  (1,  —  1,  1)  in  the 
direction  1:2:3. 

86.  Find  a  point  on  the  line  5x  +  3y  — 1=0,  3y  —  5z  — 11  =  0 
equally  distant  from  the  planes  3  x  +  3y  —  2  =  0,  4  x  +  y  +  z  +  4  =  0. 

87.  Find  the  equations  of  the  projection  of  the  line  x  -f-  y  +  z 
-2  =  0,  a-  +  2?/  +  s-2  =  0  upon  the  plane  3x  +  y  +  3z  —  1  =  0. 

88.  Find  the  length  of  the  projection  of  the  straight  line  joining  the 
points  (2,  3,  4),  (0,  —3, 1)  upon  the  straight  line  '         ='— —  =  '"  i}    ■ 

89.  Prove  that  the  plane  5  x  -f-  3  y  —  4  z  —  35  =  0  is  tangent  to 
the  sphere  (x  +  l)2  +  {y  -  2)2  +  («  -  4)2  =  50. 

90.  Find  the  center  of  the  circle  cut  from  the  sphere  x2  +  y2 
+  z2  =  49  by  the  plane  4  cc  +  6  y  +  12  *  -  49  =  0. 

91.  Find  the  equation  of  a  plane  passing  through  the  line 
x  +  3y  +  3z  +  l  =  0,  y  +  2z  +  l  =  Q  and  parallel  to  the  line 
2x  +  y-z  =  0,3x  +  2z-7  =  0. 

92.  Find  the  center  of  a  sphere  of  radius  7,  passing  through  the 
points  (2,  4,  —  4)  and  (3,  —1,-4)  and  tangent  to  the  plane  3  x  —  6  y 
+  2z  +  51=0. 

93.  What  kind  of  line  is  represented  by  the  equations  x2  +  z? 
_  4  y  =  0,  y-2  =  0? 

94.  What  kind  of  line  is  represented  by  the  equations  x2  —  9y 
-36  =  0,  x  +  5  =  0? 

95.  What  is  the  projection  of  the  curve  y2  +  z2—  6  x  =  0,  z2  =  £y 
on  the  plane  XOY? 

96.  What  is  the  projection  of  the  curve  x2  +  y2  =  a2,  y2  +  ,-2  =  a2 
on  the  plane  XOZ  ? 

97.  Find  the  projection  of  the  curve  x2  +  3  y2  —  z2  =  0,  a?2  +  y2 
—  2  a;  =  0  on  the  plane  XOZ. 

98.  Find  the  projection  of  the  curve  x2  +  2  y2  —  z2  =  1,  2  x2  —  y2 
=  8«on  the  plane  YOZ. 

99.  Show  that  the  curve  x2+  y2  =  a2,  y  =  z  is  an  ellipse.  (Rotate 
the  axes  about  OX  through  45°.) 

100.  Find  the  projections  of  the  skew  cubic  x  =  t,  y  =  t2,  z  =  £s 
on  the  coordinate  planes. 


PROBLEMS  333 

101.  Prove  that  the  projections  of  the  helix  x  =  a  cos  6,  y  =  a  sin  0, 
z  =  kd  on  the  planes  XOZ  and  YOZ  are  sine  curves,  the  width  of 
each  arch  of  which  is  kir. 

102.  What  is  the  projection  of  the  curve  x  =  e',  y  =  e~\  z  =  t  V5 
on  the  plane  XOV? 

103.  Turn  the  plane  XOZ  about  OZ  as  an  axis  through  an  angle 
of  45°,  and  show  that  the  projection  of  the  curve  x  =  e',  y  —  e~  \ 
z  =  t  V2  on  the  new  XOZ  plane  is  a  catenary. 

104.  Show  that  the  curve  x  =  tr,  y  =  2t,  z  =  t  is  a  plane  section 
of  a  parabolic  cylinder. 

105.  Prove  that  the  skew  quartic  x  =  t,  y  =  ts,  z  =  t*  is  the  inter- 
section of  an  hyperbolic  paraboloid  and  a  cylinder  of  which  the 
directrix  is  the  cubical  parabola  y  =  xs. 

106.  The  vertical  angle  of  a  cone  of  revolution  is  90°,  its  vertex 
is  at  O,  and  its  axis  coincides  with  OZ.  A  point,  starting  from  the 
vertex,  moves  in  a  spiral  path  along  the  surface  of  the  cone  so  that 
the  measure  of  the  distance  it  has  traveled  parallel  to  the  axis  of 
the  cone  is  equal  to  the  circular  measure  of  the  angle  through  which 
it  has  revolved  about  the  axis  of  the  cone.  Prove  that  the  equations 
of  its  path,  called  the  conical  helix,  are  .;•  =  t  cos  t,  y  =  t  sin  t,  z  =  t. 

107.  Show  that  the  helix  makes  a  constant  angle  with  the  elements 
of  the  cylinder  on  which  it  is  drawn. 

108.  Find  the  angle  between  the  conical  helix  x=tcos  t,  y=t8in  t} 

z  =  t  and  the  axis  of  the  cone,  for  the  point  t  =  2. 

109.  Show  that  the  angle  between  the  conical  helix  x  =  t  cos  t, 

y  =  t  sin  t,  z  =  t  and  the  element  of  the  cone  is  tan-1  — -=• 
J  V2 

110.  At  what  angle  does  the  curve  x  =  a  (1  —  cos  6),  y  =  a  sin  6, 
z  =  <i6  intersect  the  straight  line  passing  through  the  origin  and 
making  equal  angles  with  the  three  coordinate  axes  ? 

111.  Find  the  length  of  the  curve  x  =  t2,  y  —  2  t,  z  =  t  from  the 
origin  to  the  point  for  which  t  =  1. 

112.  Find  the  length  of  the  curve  x  =■  e*,  y  =  t~',  z  =  t  V2 
between  the  points  for  which  t  =  0  and  t  =  1. 

113.  Find  the  length  of  the  curve  x  =  t2 
the  origin  to  the  point  (9,  9,  6). 


334  SPACE  GEOMETRY 

114.  Find  the  length  of  the  curve  x  =  t  cos  2 1,  y=ts'm2t, 
3»  =  4i*  between  the  points  for  which  £  =  0  and  t  =  l. 

115.  Find  the  equations  of  the  tangent  line  and  the  equation  of 
the  normal  plane  to  the  curve  x  =  t~,  y  =  2  t,  z  =  t  at  the  point  for 
which  t  =  1. 

116.  Find  the  equations  of  the  tangent  line  and  the  equation  of 
the  normal  plane  to  the  curve  x  =  e*,  y  =  e~*,  »  ==  t  V2  at  the  point 
for  which  t  =  0. 

117.  Find  the  equations  of  the  tangent  line  and  the  equation  of 
the  normal  plane  to  the  curve  x  —  2  t'1  +  1,  y  =  t  —  1,  z  =  3  t3  at  the 
point  where  it  crosses  the  plane  XOZ. 

118.  Find  the  equations  of  the  tangent  line  and  the  equation  of 
the  normal  plane  to  the  conical  helix  x  =  t  cos  t,  y  =  t  sin  t,  z  =  t 

IT 

at  the  point  for  which  t  —  —  • 

119.  Find  the  equations  of  the  tangent  line  and  the  equation  of 
the  normal  plane  to  the  skew  quartic  x  =  t,  y  =  t3,  z  =  tA  at  the 
point  for  which  t  =  1. 


CHAPTER  XV 

PARTIAL  DIFFERENTIATION 

162.  Partial  derivatives.  Consider  f(x,  y~),  where  x  and  y  are 
independent  variables.  We  may,  if  we  choose,  allow  x  alone  to 
vary,  holding  y  temporarily  constant  We  thus  reduce /'(./•.  //)  to 
a  function  of  x  alone,  which  may  have  a  derivative,  defined  and 
computed  as  for  any  function  of  one  variable.  This  derivative 
is  called  the  -partial  derivative  off(x,  y)  with  respect  to  x,  and  is 

cf(  r   //) ' 
denoted  by  the  symbol    '    .  J  '     •    Thus,  by  definition, 

cf(.r,  //)  _  j  hn , f(r  +  Ax,  y)-f(x,  y) 

CX  AiiO  Ax 

Similarly,  if  ./•  is  held  constant,  f(x,  y)  becomes  temporarily  a 
function  of  //,  whose  derivative  is  called  the  partial  derivative  of 

f(x,  //)  ivith  respect  /<<  //,  denoted  by  the  symbol    '     '  '  — •    Then 

CIJ 

z/(><>/)_Umf(^y  +  ^~)-f(^y)  (2) 

cy  ^y=o  Ay 

Graphically,  if  2  =/(./-,  y)  is  represented  by  a  surface,  the  rela- 
tion between  z  and  x  when  y  is  held  constant  is  represented  by 
the  curve  of  intersection  of  the  surface  and  the  plane  y  =  const., 

and  —  is  the  slope  of  this  curve.     Also,  the  relation  between  z 

ex 

and  y  when  ./•  is  constant  is  represented  by  the  curve  of  inter- 
section of  the  surface  and  a  plane  x  =  const.,  and  —  is  the  slope 
of  this  curve. 

Thus,  in  fig.  204,  if  PQSR  represents  a  portion  of  the  surface 
z  =f(x,  y),  PQ  is  the  curve  y  =  const.,  and  PI!  is  the  curve 
x  =  const.  Let  P  be  the  point  (x,  y,  z),  and  LK=PK'=Ax, 
LM=  PM'  =  Ay. 


336 


PARTIAL  DIFFERENTIATION 


Then    LP=f(x,y),    KQ=f(x  +  Az,  y),    MR=f(z,y  +  Ay-), 

K'Q=f(x  +  Ax,y)-f(x,y),  M'Ji=f(x,  y  +  Ay)  -f(x,  y),  and 

dz      t  •     K'Q        1  *  C 

— -  =  Lim =  slope  ot  PQ, 

dx  PK' 

dz      T  .     M'B       .  ,  „ 

— -  =  Lim :  =  slope  of  PR. 

dy  PM'  l 


z 

p_____ — —^7 

R 

M 

^^lS^==^ 

N' 

0 

/^\ 

/ 

A 

I                      h 

T 

Fig.  204 
Ex.  1.    Consider  a  perfect  gas  obeying  the  law 


ct 


We  may  change 


the  temperature  while  keeping  the  pressure  unchanged.    If  At  and  Av  are 
corresponding  increments  of  t  and  v,  then 

c(t  +  A/)       ct       cAt 

Av  —  — '- = 

P  P        P 

and 


dv  _  c 
dt  ~  p 


Or  we  may  change  the  pressure  while  keeping  the  temperature  un- 
changed.   If  Ap  and  At"  are  corresponding  increments  of  p  and  v,  then 

A..-         ct  ct  =  ctAp 

p      p< 

and 


*P 


p  +  Ap 

_ct_ 

-2- 


.PAP 


So,  in  general,  if  we  have  a  function  of  any  number  of  variables 
f(x,  y, —  •  •,  g),  we  may  have  a  partial  derivative  with  respect  to 
each  of  the  variables.  These  derivatives  are  expressed  by  the  sym- 
df   of  df 


bols 


dx    dy 


••>  -T- »  or  sometimes  hyfx(x,  y, 


0»/»0*y»  •■•■»*)» 


DEKIVATIVES 


337 


•■;fg(z,  y,  •  •  •,  z).  To  compute  these  derivatives,  we  have  to 
apply  the  formulas  for  the  derivative  of  a  function  of  one  variable, 
regarding  as  constant  all  the  variables  except  the  one  with  respect 
to  which  we  differentiate. 


Ex.  2.  /=  x3  —  3  x2y  +  y3, 


Ex.  3.  /=  sin  (a2  +  y2), 


V  _ 

dx 

3x2- 

6xy, 

c.r 

2 x cos (xa 

+ 

.'/-•) 

of  _ 

-3x2 

+  3y2 
Ex. 

4 

•  /  = 

lo 

g  Vx 

2  y  cos  (x2 

+ 

/r) 

2  + 

!f 

+  z-, 

sf  = 

ex 

.  ./• 

x- 

+  .'/- 

+ 

z2' 

.'/ 

T 

+  <r 

+ 

z-z' 

?f_ 

2 

cz 

X 

+  f 

+ 

-•J 

Ex.  5.    Tn  differentiating  in  this  way  can'  must  be  taken  to  have  the 
functions  expressed  in  terms  of  the  independent  variables.    Let 


Then 


X  =  r  cos  6, 

y  =  r  sin  V. 

dx            a 
—  =  cos  a, 
dr 

dr 

dx 

0, 

84  =  reos6, 

(6 

Also,  since  r  =  Vr  +  y'2  and 


where  r  and  6  are  the  independent  variables 

x 
dr  x  n  dr 

die 


Vx2  +  y2 
x2  +  y2 


C08( 


^       Vx^ 


sin  i 
r 


< //      x2  +  y2 


(1) 


(2) 


where  x  and  ?/  are  the  independent  variables. 

It  is  to  be  emphasized  that  —  in  (1)  is  not  the  reciprocal  of  —  in  (2). 

dx  c>  to 

In  (1)  —  means  the  limit  of  the  ratio  of  the  increment  of  x  to  an  increment 

of  r  when  0  is  constant.    Graphically  (fig-  205),  OP  =  r  is  increased  by 

dx  PR 

PQ  =  Ar,  and  PR  =  Ax  is  thus  determined.    Then  —  =  Lim  — —  =  cos  6. 

cr  PQ 


338 


PA  RTIAL  DIFFERED  TIATION 


Also  —  in  (2)  means  the  limit  of  the  ratio  of  the  increment  of  r  to  that 
of  ./■  when  y  is  constant.  Graphically  (fig.  206),  OM  =  £  is  increased  by  MN  = 

PQ  =  Ax,  and  JIQ  =  A/-  is  thus  determined.    Then  —  =  Lim  — —  =  cos  6. 

dx      dr  Sx  c6     dX  PQ 

It  happens  here  that  —  =  —  But  —  in  (1)  and  —  in  (2)  are  neither  equal 

.  .  Br       dx  dd       V  J  ex       K  J  L 

nor  reciprocal. 


7 

Q 

1 

R 

\° 

0 

J 

/ 

- 

p 

Q 

<ff 

/\o 

0 

M        N 

In  cases  where   ambiguity  is   likely  to  arise   as  to  which  variable  is 

constant  in  a  partial  derivative,  the  symbol  for  the  derivative  is  sometimes 

inclosed  in  a  parenthesis  and  the  constant  variable  is  written  as  a  subscript, 

..        (dx 

thus  (  — 

\cr 


But  it 


may 


be  shown 


163.  Higher  partial  derivatives.  The  partial  derivatives  of 
f(x,  if)  are  themselves  functions  of  x  and  y  which  may  have 
partial  derivatives,  called  the  second  partial  derivatives  of/(:r,  y). 

L(dl),  ±(dl\  ndi\  ^(d-t 
dx\dxj    dy\dx)    dx\dy)    dy\dy/ 

that  the  order  of  differentiation  with  respect  to  x  and  y  is  imma- 
terial when  the  functions  and  their  derivative  fulfill  the  ordinary 
conditions  as  to  continuity,  so  that  the  second  partial  derivatives 
are  three  in  number,  expressed  by  the  symbols 


They 


)x\dx)      dx2     Jx 


dx 

d_fdf 
dx\cy 


d_(df_ 

dy\dx 


dxby 


Jxyi 


dy\dy>      dy 


rt 


2       Jyi 


INCREMENT  AND  DIFFERENTIAL  339 


Similarly,   the   third  partial  derivatives  of  f(x. 
number,  namely, 

,  y)  are 

four   in 

urn, 

dx\cx-J 

of 
cxs' 

a  /ay\ 

dyW)~ 

=  g  /  gy  \ 

a2  /an 

=  da?\dy) 

.  JZ_, 

drdy 

d  /dj\ 
dx\cy'y 

_g/gyv 

dy\dxdy) 

df\dx) 

ay 

cxcy'1 

a;' + ''/* 

So,  in  general,   ^  aiS      signifies  the  result  of  differentiating 

f (■>'■<  >/)  p  times  with  respect  to  x  and  q  times  with  respect  to 
?/,  the  order  of  differentiating  being  immaterial. 

The  extension  to  any  number  of  variable's  is  obvious. 

164.  Increment  and  differential  of  a  function  of  two  variables. 
Consider  z=f(x,  ?/),  and  let  x  and  y  be  given  any  increments 
Ax  and  Ay.    Then  z  takes  an  increment  A.?,  where 

A2  =/(.,•  +  Ax,  n  +  Ay)  -/(./■.  // ).  (1) 

In  fig.  204,       A7>*  =  2'  =/<  ./•  +  A/-,  v/  +  Ay  ) 
and  J!FS=z' -z  =  Az.  (2) 

If  :r  and  y  are  independent  variables.  A./'  and  Ay  are  also 
independent.  Thus  the  position  of  S  in  fig.  201  depends  upon 
the  choice  of  LK  and  LM,  which  can  be  taken  at  pleasure. 

The  function  z  is  called  a  continuous  function  of  x  and  y  if  A.z 
approaches  zero  as  a  limit  when  A./-  and  Ay  approach  zero  as  a 
limit  in  any  manner  whatever. 

Tims,  in  fig.  201,  if  2  is  a  continuous  function  of  x  and  y, 
the  point  S  will  approach  the  point  P  as  LK  and  LM  approach 
zero,  no  matter  what  curve  the  point  N  traces  on  the  plane 
XOY  or  the  point  S  on  the  surface. 

We  shall  assume  that  z  and  its  derivatives  are  continuous 
functions. 


340  PARTIAL  DIFFERENTIATION 

The  expression  for  Az  may  be  modified  as  follows : 
The  line  N'S  may  be  separated  into  two  portions  by  drawing 
from  Q  a  line  parallel  to  K'N  meeting  NS  in  N".    Then 

N'S  =  N'N"  +  N"S=  K'  Q  +  N"  S.  (3) 

The    line   K'Q  is  connected  with    the    slope    of  PQ  by  the 

relation  K,  ~  „ 

Lim  - — r  =  slope  of  PQ  =  ~, 
PA'  ex 

the  limit  being  taken  as  PK'  =  Ax  approaches  zero. 

K'Q      dz 

Hence  — -  = he,, 

PK'      dx       * 

where  ex  approaches  zero  as  Ax  approaches  zero,  so  that 

a"M!+%>a*-  (4) 

Also  the  line  N" S  is  connected  with  the  slope  of  QS  by  the 
relation  N,,  „ 

Lim  — ~—  =  slope  of  QS, 

the  limit  being  taken  as   QN"  =  Ay  approaches  zero.     But  as 

Ax  =  0,  the  curve  QS  approaches  the  curve  PR.    Hence  we  are 

justified  in  saying 

N"  S  dz 

Lim =  slope  of  PR  =  — , 

QN"  F  dy 

the  limit  being  taken  as  both  Ax  and  Ay  approach  zero. 

N"S      dz 

Hence  -= |-e„, 

QN"      dy       2 

where  e2  approaches  zero  as  Ax  and  Ay  approach  zero,  so  that 

**«-(!+■,,)  ah  (5) 

since  QN"  =  PM'  =  Ay. 

Substituting  from  (4)  and  (5)  in  (3)  and  then  in  (2),  we 
have  j,  o 

Az  =  ■£  Ax  +  ^  Ay  +  exAx  +  e^Ay.  (6) 


DIFFERENTIAL  341 

In  a  manner  analogous  to  the  procedure  in  the  case  of  a 
function  of  one  variable  (§  77),  we  separate  from  the  incre- 
ment the  terms  eJAz  +  e„Ay,  call  the  remaining  terms  the 
total  differential  of  the  function,  and  denote  them  by  dz.  The 
differentials  of  the  independent  variables  are  taken  equal  to 
the  increments,  as  in  §  77.  Thus,  we  have  by  definition,  when 
z  is  a  function  of  two  independent  variables  x  and  y, 

,       dz  ,        dz  ,  ,„^ 

dz  =  —  dx  +  —  dy.  (7) 

dx  Cy  K  y 

In  (7)  dx  and  dy  may  be  given  any  values  whatever.  If, 
in  particular,  we  place  either  one  equal  to.  zero,  we  have  the 
partial  differentials,  indicated  by  dxz  and  dyz.    Thus 

dxz  =  —  dx,  d  z  =  —  dy. 

ex  "       dy 

A  partial  differential  expresses  approximately  the  change  in 
the  function  caused  by  a  change  in  one  of  the  independent 
variables ;  the  total  differential  expresses  approximately  the 
change  in  the  function  caused  by  changes  in  all  the  inde- 
pendent variables.  It  appears  from  (7)  that  the  total  differen- 
tial is  the  sum  of  the  partial  differentials. 

Ex.    The  period  of  a  simple  pendulum  with  small  oscillations  is 


•Wj- 


T 

U 

.  \    TV -I 

J  y.o 

Let  /  =  100  cm.  with    a   possible    error  of    },  mm.   in    measuring  and 
T  =  2  sec.  with  a  possible  error  of  T^ff  sec.  in  measuring.    Then  dl  =  ±  JG 

Also  dg  =  ±£dl.-%£dT, 

and  we  obtain  the  largest  possible  error  in  g  by  taking  dl  and  dT  of  oppo- 


10.36. 


site  signs, 

say  J/  = 

5(j»  dl 

i 

[TJTJ- 

Then                         dg  - 
The  ratio  of  error  is 

■s* 

7T2  = 

:  1.05  7T 

dg  = 

.'11-0 

I       " 

dT  _ 

T  " 

.0005  +  .01 

.0105  =  1.05%. 


342  PARTIAL  DIFFERENTIATION 

165.  Extension  to  three  or  more  variables.  The  results  of  the 
previous  article  may  be  extended  to  the  cases  of  three  or  more 
independent  variables  by  reasoning  which  is  essentially  that  just 
employed,  without  the  geometric  interpretation,  which  is  now 
impossible.    For  example,  consider 

u=f{x,  y,  z~).  (1) 

Let  x,  y,  z  be  given  increments  Ax,  Ay,  Az,  and  let 
u'  =f(x  -{-Ax,  y  +  Ay,  z  +  Az). 
Then  Am  =  u'  —  u. 

For  convenience,  introduce  new  functions 

ux  =f(z  +  Ax,  y  +  Ay,  z), 

u2  =f(.x  +  A^'  y>  g> 

(2) 


Then 

Au  —  U1  —  Mj  +  Mj  —  U2  +  U2  —  U. 

Now 

S  T  .     u0  —  u      du        ,            u„  —  u      du  , 
Lim    \       =      ,    whence     \       =      +  e, 

ax=0     Aa;        dx                       Ax        ox 

x  .     w.  —  u0      du„       ,            u,  —  w„      dw,,  , 
Lim  -±- — -  =  — - ,  whence  -±- — -  =  — -  -+-  e  , 
a^o      Ay          dy                        Ay          dy 

T  .     w'  —  u,      du,        ,            m'  —  u,      du, 
Lim  — - — i  =  — i ,  whence  — - — -  =  -^  +  e  , 

Az  =  0       A«             02                               Az             cs 

that 

w —  u  =  —  As  +  eAx, 

2              eb              * 

w1-M„  =  -JAy  +  eAy, 
1        -      dy 

"'-"■=  &A2+e^- 

But  —  and  —  approach  —  and  —  as  Ax  and  Ay  approach 

dy  dz  dy  dz 

,  du0      du                ^    du,      du  ,          TT            ,ON 

zero   so  that  — *-  =  —  +  e±  and  — ±  =  —  +  e_.    Hence   (2)   may 

,           tJJ  dy      dy       *             dz       dz 

be  written  ■*        ^ 

Aw  =  —  Ax  +  —  Ay  +  —  Az  +  e,Ax  +  e^Ay  +  e^Az,        (3) 
dx  dy  dz 

where       4  =  e2  +  e4     an(^  ,  es  =  es  +  e5- 


DIRECTIONAL  DERIVATIVE  343 

Then  du  is  the  part  of  this  expression  which  does  not  contain 
ev  e'2,  or  e'3,  with  the  increments  of  the  independent  variables 
replaced  as  usual  by  their  differentials.    That  is, 

du  =  —  ate  4 du  -\ dz.  (4) 

OX  C I i    '         cz 

The  extension  to  more  variables  is  obvious. 

166.  Directional  derivative  of  a  function  of  two  variables.   The 

result  of  §  164  may  be  used  to  find  the  slope  of  PS  (fig.  204), 

which  is  a  curve  cut   out  of  the   surface  z  —f(z,  y)  by  any 

plane  through  LP.    Draw  the  lines   FN1  and  LN  as  shown  in 

the  figure,  and  let  r  ,r      „,rt      . 

ft  LN  =  PX'  =  Ar, 

where  r  is  the  distance  measured  from  some  point  on  the  line 
LN  produced.  Denote  by  6  the  angle  KLX=  h"  /'X'.  which 
is  equal  to  the  angle  made  by  the  ['lane  of  PS  with  the  plane 
ZOX.    Then    __       .  _  „       . 

T.k         A  >•  I    M        A  a 

6, 


LK      Ax  n      LM      Ay 

=  cos  0, =  — -  =  sin 


LX      A/-  LN      A/- 

and  the  slope  of  PS=  Lim  '    -  =  Lim  -—  =  — 1  the  limit  being 
PN  A/-      dr 

taken  as  S  approaches  P  along  PS. 

From  (6),  §  164, 

A*      cz  Ax      cz  Ay  Ax         A// 

—  = 1 -  +  e he.— - 

A/-      fa;  A/-      cy  Ar        J  Ar        "  Ar 

=  —  cos  9  +  —  sin  0  4-  e  cos  #  4-  e  sin  0. 
ax-  e  y 

Taking  the  limit,  we  have 

'lz  =  c-l  cos  #  +  ££  sin  #  =  siope  of  P& 
ar      ex  £*/ 

Now  —  measures  the  rate  of  change  of  z  in  the  direction 

dz  dz 

LK,  —  the  rate  of  change  in  the  direction  LM.  and  —  the  rate 

C'J         .  .   dV.        dz 

of  change  in  the  general  direction   LN.    The  derivative  —   is 

called  the  directional  derivative  in   the   direction  of  /\ 


344  PARTIAL  DIFFERENTIATION" 

Ex.    The  temperature  u  at  any  point  of  a  plate  is  given  by  the  formula 

u  —  — -  ■    Find  at  the  point  (2,  3)  the  rate  of  change  of  temperature 

in  the  direction  making  an  angle  of  30°  with  OX. 

w.i.™,  8u-       2x        du-       2n 


dx 

(x*  + 

fy> 

ty 

(x2  + 

ff 

d  at  the  point  in  question 

du 

dx~ 

4 
169' 

du 

6 
169' 

Hence  the  required  i 

rate  of  change  is 

du 

4 

= cos  30° 

169 

-  -5-  sin  30° 
169 

2V3 

+  3 

-  .038. 

169 

167.  Total  derivative  of  z  with  respect  to  x.  In  fig.  204,  let 
the  point  S  approach  the  point  P  along  any  curve  whatever  on 
the  surface,  and  not  along  the  curve  PS,  as  in  §  166.  Then  the 
point  N  describes  a  curve  on  the  plane  XO  Y,  the  equation  of 
which  may  be  taken  as  y  =  $  (:r),  and 

|  =  f(*)  (D 

is  the  slope  of  the  curve  described  by  A7". 

During  this  motion  of  the  point  S,  z  is  a  function  of  x ;  since 
it  is  in  general  a  function  of  x  and  y,  and  y  is  a  function  of  x. 
Hence  z  has  a  derivative  with  respect  to  x  and 

dz      T  .     Az 

—  =  Lim 

ax  Ax 

Dividing  the  expression  for  Az  as  given  in  (6),  §  164,  by 
Ax,  and  taking  the  limit,  we  have 

dz  =  fa  +  dzdy^ 
dx      dx      dy  dx 

This  is  the  total  derivative  of  z  with  respect  to  x. 


TANGENT  PLANE  345 

This  result  (2)  has  an  important  application  when  the  curve 
along  which  S  moves  is  on  the  plane  XO  Y.  For  then  z  =  0,  a 
constant,  and,  from  (2), 

f  +  TT  =  0'  (3) 

cx      cydx 

where  -j-  is  the  slope  of  this  curve,  as  shown  in  (1).   We  express 
this  result  in  the  following  theorem: 

1.  The  value  of -j-  may  he  found  from  the  equation 

/(*,£) =o 

by  the  formula  ¥.  +  dl^l  =  0. 

Cx      Cy  ax 

Again,  let  z  be  defined  as  an  implicit  function  of  x  and  y  by 
the  equation  F^  ^  ^  _  0> 

If  we  hold  y  constant  temporarily,  the  case  reduces  to  the  one 

dz 
discussed  in  theorem  1,  with  z  in  place  of  y  and  —  in  place 

7  OX 

of  -j—     Similarly,  if  we  hold  x  temporarily  constant,  we  get 
theorem  1  with  change  of  letters.     Hence : 

2.  The  values  of—  <<n<l     '  may  be  found  from  tin-  equation 

dy 

Ffr  y,  g)=0 

by  the  formulas  - — (-—-—-  =  0, 

cx      cz  ox 

cF     c_Fdz__ 
dy       cz  cy 

168.  The  tangent  plane.  In  fig.  204,  let  P  be  given  the  fixed 
coordinates  (x^  yv  zx).  The  tangent  line  to  PQ  in  the  plane 
y  =  y1  is,  by  §  76,  /g  , 

-V  (*),<—'*>■  (1) 

and  the  tangent  line  to  PR  in  the  plane  x  —  x1  is 


346 


P  APT  I AL  1)  I F  F  E  R  EN  T I A  TION 


Both  of  these  lines  lie  in,  and  hence  determine,  the  plane  of 
which  the  equation  is 

for  this  equation  reduces  to  (1)  when  y  =  y     and  reduces  to 
(2)  when  x  =  %x. 

This  plane  is  called  the  tangent  plane  to  the  surface  at  the  point 

We  shall  prove  that  the  plane  (3)  contains  all  tangent  lines  to 
the  surface  z=f(x,  if)  which  pass  through  P. 

The  line  through  the  two  points  P  and  S  has  the  direction 
Ax :  Ay :  As.    Its  equations  are  therefore 


x  -  xx  =y-  yx 

Ax  Ay 


Az 


A- 
Ax 


(x-x\ 


(4) 


(5) 


As  the  point  S  approaches  the  point  P,  the  line  (5)  ap- 
proaches as  a  limit  a  tangent  line  at  P,  and  the  equations 
of  this  tangent  are 


'dz      dz  dy 


ex      cy  dx/i  l 


(6) 


An  easy  combination  of  these  equations  gives  (3)  as  an  equation 
satisfied  by  any  tangent  line.     Hence  the  theorem  is  proved. 
If  the  equation  of  the  surface  is  given  in  the  form 


F(x,  y,  Z)=0, 


(?) 


TANGENT  PLANE  347 

the  equation  of  the  tangent  plane  may  be  found  without  solving 
for  z.    For,  from  theorem  2,  §  107, 

d_F 

dz  _      dx 
dx~~d£ 

~dz~ 

cF 

dy _     dF 

cz 

Substituting  these  values  in   (3)  and  making  a  few  simple 
changes,  we  have  as  the  equation  of  the  tangent  plane, 

(fKH^M+(f),H)-  - 

The  straight  line  perpendicular  to  the  tangent  plane  at   the 
point  of  eontaet  is  the  normal  to  the  surface.    Its  equations  are 

(9) 
(10) 


x  —  X 

y  -  //,    z  -  *, 

(-)    _1 

y  -  Hx     z  - z, 

\exl     UyA    \d*h 


Ex.  1.    Find  the  tangent  plane  and  the  normal  line  t<>  the  paraboloid 

-  =  a.r'1  +  by2. 

dz  dz 

Here  —  =  2  ax  and        =2  hi/.    Hence  the  tangent  plane  is 
dx  dij 

2  ax,  (x  -  x, )  +  2  by,  (//  -  y, )  -  (z  -  zx)  =  0, 

or  2  ax,x  +  2  byxy  -  2  axf  -  2  by?  -z  +  z,=  0. 

But  since  2  ax,  +  2  by,  =  2  ~r  this  may  be  written 
2  «xrr  +  2  6y^  -z  —  z1  =  0. 
x  ~  xi  _  V  ~  V\  _  g-8!  . 


The  normal  is  —i — -* 


2  oxx         2  by, 


348  PARTIAL  DIFFERENTIATION 

Ex.  2.    Find  the  tangent  plane  and  the  normal  line  to  the  ellipsoid 


TT  dF      2  x      dF      2  y       dF 

Here 


dx        a'2       dy        0'z         dz        c2 
Hence  the  tangent  plane  is 

o  r  o  w  o  - 

a2       i2       c2 

since  -L  +  ^L  4.  II  =  l, 

a2       ft2       c2 

The  normal  line  is       _L  =  2 1!  =  I 11 . 

£1  2/i  £1 

a2  b"  c2 

If  a  curve  is  denned  as  the  intersection  of  two  surfaces  by 
the  equations 

f(x,y,z)=0,  F(x,y,z)=0, 

its  tangent  line  is  evidently  the  intersection  of  the  two  tangent 
planes  to  these  surfaces.  The  equations  of  the  tangent  line  are 
therefore  two  equations  of  the  form  (8).  The  direction  cosines 
of  the  tangent  line  can  be  found  by  the  method  of  §  156.  The 
normal  plane  may  be  found  by  the  method  of  §  151. 

169.  Maxima  and  minima.  In  order  that  the  function  f(x,  y) 
shall  have  a  maximum  or  a  minimum  value  for  x  =  x%,  y  =  yx, 
it  is  necessary,  but  not  sufficient,  that  the  tangent  plane  to  the 
surface  z  =f(x,  y~)  at  the  point  (.^,  yx,  z^)  should  be  parallel 

to  the  plane  XOY.    This  occurs  when  (y)=  0,  (f-)=°-    These 

are  therefore  necessary  conditions  for  a  maximum  or  a  mini- 
mum, and  in  case  the  existence  of  a  maximum  or  a  minimum 
is  known  from  the  nature  of  the  problem,  it  may  be  located 
by  solving  these  equations. 

Ex.  It  is  required  to  consflrnct  ont  of  a  given  amount  of  material  a  cistern 
in  the  form  of  a  rectangular  parallelepiped  open  at  the  top.  Required  the 
dimensions  in  order  that  the  capacity  may  be  a  maximum,  if  no  allowance 
is  made  for  thickness  of  the  material  or  waste  in  construction. 


EXACT  DIFFERENTIALS  349 

Let  x,  y,  z  be  the  length,  the  breadth,  and  the  height  respectively.  Then 
the  superficial  area  is  xy  +  2  xz  +  2  yz,  which  may  be  placed  equal  to  the 
given  amount  of  material,  a.    If  v  is  the  capacity  of  the  cistern, 

axy  —  x2t/2 

V  =  XlIZ  =  - —  • 

2(x  +  y) 

rr\  &°  _  (a  ~  -  xll  ~  x2)  V2  ?v  _  (a  ~  2  xy  —  y")  x2 

d~x~  2(x  +  y)2  JJj~         2(x  +  y)2 

For  the  maximum  these  must  be  zero,  and  since  it  is  not  admissible  to 
have  x  =  0,  y  =  0,  we  have  to  solve  the  equations 

a  —  2  xy  —  x2  =  0, 
a  —  2  xy  -  y2  =  0, 

1      fa 

,<V;- 

Consequently,  if  there  is  a  maximum  capacity,  it  must  In-  for  these  dimen- 
sions. It  is  very  evident  that  a  maximum  does  exist;  hence  the  problem 
is  solved. 


which  have  for  the  only  positive  solutions  x  =  y  =  -%/- j  whence  z 


More  generally,  if  a  function  of  three  or  more  independent 
variables  has  a  maximum  or  minimum  when  all  the  variables 
change  in  any  way,  it  must  have  a  maximum  or  minimum  when 
each  changes  alone.  Therefore,  iff(x,  y,  z)  has  a  maximum  or 
a  minimum,  it  is  necessary,  by  §  89,  that 

Sf-ov      Z-o,      2=0. 

ex  cy  cz 

170.  Exact  differentials.  We  have  seen  that  if  z  =f(x,  y),  then 

,        dz  ,        cz  ,  H^ 

When  the  function  f(.r,  y)  is  known,  the  partial  derivatives  — 

~  dx 

and  —  may  be  found,  and   the  second   member  of  (1)  is  of 

the  form  Mdx+Ndy,  (2) 

where  M  and  JV  are  functions  of  x  and  >/.  In  §  164,  (1)  was 
called  a  total  differential ;  it  will  now  be  called  an  exact  differ- 
ential, to  emphasize  the  fact  that  it  may  be  exactly  obtained 
by  differentiation. 


350  PARTIAL  DIFFERENTIATION 

Now  expressions  of  the  form  (2)  arise  in  practice  by  other 
methods  than  by  differentiation,  or  they  may  be  written  down 
at  pleasure.  For  example,  we  may  write  arbitrarily  the  two 
following  expressions :' 

(4  x»  -  2  ,-/)  dx  +  (4  f  -  2  xy)  dy,  (3) 

(a?+xy)dx  +  y*dy.  (4) 

It  is  important,  therefore,  to  know  whether  an  expression 
of  the  form  (2)  is  always  exact ;  that  is,  whether  it  is  always 
possible  to  find  z=f(x,  y)  so  that  (2)  is  equivalent  to  (1). 

In  discussing  this  question  we  note  first  that  if  (2)  is 
equivalent  to  (1),  we  must  have 

—  =  M,         —  =  N,  (5) 

ox  dy 

d'2z        dM     dN 
whence  t-^-  =  -—  =  -—. 

oxcy       dy       ex 

Hence,  if  Mdx  +  Ndy  is  an  exact  differential,  it  is  necessary 

that  d_M_d_N 

dy  ~  dx  '  ^  ' 

From  this   it  appears   that  (4)  is  not  an  exact  differential, 

since  —  =  x  and  —  =  0.    On  the  other  hand,  (3)  may  possibly 

dM  .  .  dN         , 

be  exact,  since  — —  =  —  4  xy  and  — —  =  —  4  xy. 
dy  dx 

Let  us  now  assume  that  the  condition  (6)  is  met,  and  try  to 
find  z.  We  may  integrate  the  first  equation  of  (5)  consider- 
ing y  as  a  constant.  The  constant  of  integration  then  possibly 
contains  y  and  must  be  expressed  as  a  function  of  y.    Then 


■/ 


Mdx  +  <f>(y}.  (7) 


Substituting  this  in  the  second  equation  of  f5),  we  have 
yCMdx  +  V<iy-)=N, 

4f(jj)=N~fMte.  (8) 


EXACT  DIFFERENTIALS  351 

By  hypothesis  the  first  member  of  (8)  does  not  contain  x. 
Hence  the  second  member  of  (8)  must  be  free  from  x  or  the 
work  cannot  go  on.  Now  the  condition  that  an  expression  shall 
be  free  from  x  is  that  its  derivative  with  respect  to  x  shall  be 
zero.    Hence,  from  (8),  we  must  have 

d*-fMdx  =  0.  (9) 


But 

ex 


dx       <■'<!!, 


tyj  <!l\''J  J       CU 


The  condition  (9)  is  then  simply  (<!),  which  is  fulfilled  by 
hypothesis. 

From  (8),  the  value  of  <£(//)  can  now  be  found  and  substi- 
tuted in  (7).     The  value  of  z  is  thus  found. 

We  have  accordingly  the  following  theorem,  the  converse  of 
the  one  stated  above. 

Tf  —  =  -^—t  the  expression  Hfdx-\- Ndy  is  an  exact  differential 

cy       ex 

dz,  and  i\r—^_        v— — 

dx  dy 

The  process  of  finding  z  is  illustrated  in  Kxs.  1  and  '2.  Ex.  ;,< 
shows  how  the  process  fails  if  it  is  wrongly  applied  to  an 
expression  which  is  not  an  exact   differential. 

Ex.  l.   (4  x*  -  2 ./•//-)</./•  +  ( I  y*  -  2  x*y)dy. 

Here  — —  =—  4 xy  —  —    ■    Hence  the  expression  is  equal  to  dz,  and 
By  dx 

^■  =  4x*-2xf,  (1) 

[y*-2x*y.  (2) 

Integrating  (1)  with  respect  to  x,  we  have 

z  =  x*-xY+f(y).  (3) 

Substituting  in  (2),  we  have 

-2x*y+f(y)  =  ±tf-2xhn 

whence  f  (y)  =  4  ys, 

and  /{,/)  =  yi+C. 

Substituting  in  (1),  we  have    z  —  x*  —  x2y-  +  y*  +  C. 


352  PARTIAL  DIFFERENTIATION 

Ex.  2.    (I JL=) dx  +     _L_ dy. 

V      x  Vf  -  xV  Vf  -  x2 

dM  x  dN 

Here   — —  = - — The    expression   is   therefore    an    exact 

Sy      ^2  _  x*)l      dx 

differential   dz,  and  a        , 

(1) 

(2) 


dx         X         x-VyZ-x* 

dz  1 


dV      V?/2  -  x2 
Integrating  (2)  with  respect  to  y,  we  have 


z  =  log  (y  +  V?/2  -  x2)  +  f(x).  (3) 

Substituting  in  (1),  we  have 

~x  +f(x)  =  1 y      . 

V?/2  -  x2  {y  +  Vf  -  x2)  *       a-  y[y~*  _  X2  ' 

whence  f  (x)  =  0, 

and  f(x)  =  C. 

Substituting  in  (3),  we  have     z  =  log(y  +  V#2  —  x2)  +  C. 

Ex.  3.  (x2  +  xy)  dx  +  tfdy. 

Here  =  x,    —  =  0, 

dy  dx 

and  the  expre  ssion  is  not  exact.    If  one  wrongly  put 
and  integrated  (1)  with  respect  to  x,  he  would  have 


Substituting  in  (2),  he  would  have 


2  +  /'(.'/)  =  >/3 


x2 
whence  /'(#)  =  if  —  —• 

But /'(?/)  should  be  a  function  of  ?/  alone,  and  the  last  equation  is  absurd. 
Equations  (1)  and  (2)  are  therefore  false. 


LINE  INTEGRALS  353 

171.  Line   integrals.     The   expression  Mdx+Ndy  occurs  in 
certain  problems  involving  the  limit  of  a  sum  as  follows : 

Let  C  (fig.  207)  be  any  curve  in  the  plane  XOY  connecting 
the  two  points  L  and  K,  and  let  M  and  N  be  two  functions  of 
x  and  y  which  are  one-valued  and  continuous  for  all  points 
on  C.  Let  C  be  divided  into  n  segments  by  the  points  Pv  i£, 
Pz,  -  •  -,  Pn_v  and  let  Ax  be  the  pro- 
jection of  one  of  these  segments  on 
OX  and  Ay  its  projection  on  OY. 
That  is,  Ax  =  xi+1  —  x„  Ay  =  yi  +  1  —  yn 
where  the  values  of  Ax  and  Ay  arc 
not  necessarily  the  same  for  all 
values  of  i.  Let  the  value  of  M  for 
each  of  the  n  points  L,  Pv  P,,  •  •  -,       °  pIG  2o7 

Pl_1  be  multiplied  by  the  correspond- 
ing value  of  Ax,  and  the  value  of  N  for  the  same  point  by 
the  corresponding  value  of  Ay,  and  let  the  sum 

i=n-l 

%[M(x„  y{)Ax+N(z{,  yJAy] 

be  formed. 

The  limit  of  this  sum  as  n  increases  without   limit  and  Ax 
and  Ay  approach  zero  as  a  limit  is  denoted  by 


/ 

J(C) 


(  Mdx+Ndy), 

'(C) 

and  is  called  a  line  integral  along  the  curve  C.    The  point  K  may 
coincide  with  the  point  L,  thus  making  C  a  closed  curve. 

Ex.  1.  Work.  Let  us  assume  that  at  every  point  of  the  plane  there  acts 
a  force  which  varies  from  point  to  point  in  magnitude  and  direction.  We 
wish  to  find  the  work  done  on  a  particle  moving  from  L  to  A'  along  the 
curve  C.  Let  C  be  divided  into  segments,  each  of  which  is  denoted  by  As 
and  one  of  which  is  represented  in  fig.  208  by  PQ.  Let  F  be  the  force 
acting  at  P,  PR  the  direction  in  which  it  acts,  PT  the  tangent  to  C  at  P, 
and  6  the  angle  PPT.  Then  the  component  of  F  in  the  direction  PT  is 
F  cos  6,  and  the  work  done  on  a  particle  moving  from  P  to  Q  is  F  cos  6  As, 
except  for  infinitesimals  of  higher  order.  The  work  W  done  in  moving 
the  particle  along  C  is,  therefore, 

TF  =  Lim  V  Fcoa  6As=   f  Fcos  Ods. 


354 


1  'A  RTIAL   1 )  1 F  FE I ',  EN  T I ATION 


Now  letabe  the  angle  between  PR  and  OX,  and  4>  the  angle  between  P77 
and  ".V.    Then  0  =  <jt  —  a  and  cos  6  —  cos  «£  cos  a  +  sin  <f> -sin  a.    Therefore 


X(7''  cos  <£  eo.s  a  +  7'7  sin  <f>  sin  a)  ds. 


But  /•' cos  a  is  the  component  of  force  parallel  to  OX  and  is  usually 
denoted  by  A'.  Also  F  sin  a  is  the  component  of  force  parallel  to  ()Y 
ami  is  usually  denoted  by  )'.  Moreover, 
COSffids  =  dx  and  fiin<jn/s  =  dy  (§  !»1  ). 
Hence   we  have,  finally, 


=   f   (Xdx  +  Ydy). 


Ex.  2.  Heat.  Consider  a  substance  in 
a  given  state  of  pressure  p,  volume  v,  and 
temperature  t.  Then  p,  v,  t  are  connected 
by  a  relation  f(p,  v,  t)  =  0,  so  that  any 
two  of  them  may  be  taken  as  independent 
variables.  We  shall  take  t  and  v  as  the 
independent  variables  and  shall  therefore  work  on  the   (/,   r)  plane. 

Now  if  Q  is  the  amount  of  heat  in  the  substance  and  an  amount  dQ  is 
added,  there  result  changes  dv  and  dt  in  v  and  t  respectively,  and,  except 
for  infinitesimals  of  higher  order, 

dQ  =  Mdt  +  Ndv. 

Hence  the  total  amount  of  heat  introduced  into  the  substance  by  a 
variation  of  its  state  indicated  by  the  curve  C  is 


Q=    f     (Mdt  +  Ndv). 

d(C) 


Ex.  3.  A  rea.  Consider  a  closed  curve  C  (fig.  200)  tangent  to  the 
straight  lines  x  =  a,  x  =  b,  y  =  d,  and  y  =  e,  and  of  such  shape  that  a 
straight  line  parallel  to  either  of  the  coordinate  axes  intersects  it  in 
not  more  than  two  points.  Let  the 
ordinate  through  any  point  M  inter- 
sect C  in  Px  and  P2,  where  MPX  =  i/x 
and  MP„  =  ?/„.  Then,  if  A  is  the  area 
inclosed  by  the  curve, 

A  =  f  !h',x  -  f  Vidx 

=  —  f  !l-i<lx  -  f  !li,lx 

=  -  f  ydx, 

the  last  integral  being  taken  around  C  in  a  direction  opposite  to  the 
motion  of  the  hands  of  a  clock. 


: 

E 

ySfT             ^\ 

A 

1 

N 

Wi 

,    ^ 

1          ^^ 

L> 

0 

j 

1  Jl 

I                        J 
Fig.  209 

3 

LINE  INTEGRALS  855 

Similarly,  if  the  line  NQa  intersects  C  in  Qx  and  Q.2,  where  iVQj  =  x, 
and  X'l,  —  ./;.„  we  have 


l=X>-X> 

./(C) 


the  last  integral  being-  taken  also  in  the  direction  opposite  to  the  motion 
of  the  hands  of  a  clock.    By  adding  the  two  values  of  .1,  we  have 

2.1  =   f   (-ydx  +  xdy). 

■   , 

If  we  apply  this  to  find  the  area  of  an  ellipse,  we  may  take  .;•  =  a  cos  <f>, 
y  =  h  sin  </>  (§  54).    Then  .  I  =  \  C^abdtft  =  irab. 

If  the  equation  of  the  curve  C  is  known,  the  line  integral 
may  be  reduced  to  a  definite  integral  in  one  variable.  In 
general,  the  value  of  the  line  integral  depends  upon  the  curve 
C  and  not  merely  on  the  position  of  the  points  L  and  A".  This 
is  illustrated  in  Ex.  4.  If,  however,  Mdx  +  Ndy  is  an  exact 
differential  rfz,  we  shall   have 


^(Mdx  +  Ndy)=pdz  =  z^ 


where  z  and  z  are  the  values  of  z  at  the  points  A  and  K.  This 
result  is,  in  general,  independent  of  the  curve  <\  though  special 
consideration  may  be  necessary  if  z  may  take  more  than  one 
value  at   A  or   A". 

The  integral  of  an  exact  differential  taken  around  a  closed 
path  is,  in  general,  zero ;  while  the  line  integrals  of  other  dif- 
ferentials around  a  closed  path  are  not  zero. 

(ydx  -  xdy). 

(0,  0) 

Let  us  first  integrate  along  a  straight  line  connecting  0  and  Px  (fig.  210). 

The  equation  of  the  line  is  y  =  —  r,  and  therefore  along  this  line  y  dx  —  x  dy  =  0, 

xi 
and  hence  the  value  of  the  integral  is  zero. 


356 


PARTIAL  D I F  F  EE EN  TI ATION 


Next,  let  us  integrate  along  a  parabola  connecting  O  and  Pv  the  equation 

V? 
of  which  is  y2  —  —  x.    Along  this  parabola 


C0**- «-5fer^*-i 


■'u'/i- 


Next,  let  us  integrate  along  a  path  consisting  of  the  two  straight 
lines  OMx  and  MXPV  Along  OMv  y  =  0  and  dy  =  0 ;  and  along  M1PV 
x  —  xl    and    dx  =  0.      Hence    the    line    integral 

reduces  to  .—  f    lx1dy  =  —x1yv 
J  o 

Finally,  let  us  integrate  along  a  path  consist- 
ing of  the  straight  lines  ON1  and  N1PV  Along 
ONv  x  =  0  and  dx  =  0  ;  and  along  N1PV  y  —  yx 
and  e/y  =  0.  Therefore  the  line  integral  reduces 
t0  fXly1dx  =  x1yv 

"0 


Ex.  5. 


Here 


r 


vd  —  ydx  +  xdy 


Fig.  210 


Therefore 


I, 


(x0,  y0) 


ydx  +  xdy 
x2  +  y2 
ydx  +  xdy 


(1  ( tan- 


!■" 


cz  +  y1 


r>dd=ex-< 


If  the  curve  C  does  not  pass  around  O,  Qx  will  be  the  angle  shown  in 
the  figure  (fig.  211).    If,  however,  C  is  drawn  around  the  origin,  the  final 
value  of  6  is  2  ir  +  6V  and  the  value  of 
the  integral  is  2ir  +  61—  0O. 

The  value  of  this  integral  around  a 
closed  curve  is  zero  if  the  curve  does  not 
inclose  the  origin,  and  is  2  it  if  the  curve 
winds  around  the  origin  once  in  the  posi- 
tive direction. 

Ex.  6.  Work.  If  X  and  Y  are  compo- 
nents of   force  in  a   field  of    force,  and 

— —  =  — -  >  then  the  work  done  in  moving 

a   particle    between   two    points   is   inde- 
pendent of  the  path   along  which  it  is  moved,  and  the  work  done  on 
a  particle  moving  around  a  closed  curve  is  zero.     Also  there  exists  a 
function    <£,    called   a    force    function,   the    derivatives    of    which    with 
respect  to  x  and  y  give  the  components  of  force  parallel  to  the  axes 


COMPOSITE  FUNCTIONS  357 

of  x  and  y.  Such  a  force  as  this  is  called  a  conservative  force.  Examples 
are  the  force  of  gravity  and  forces  which  are  a  function  of  the  distance 
from  a  fixed  point  and  directed  along  straight  lines  passing  through 
that  point. 

If  the  components  of  force  X  and  Y  in  a  field  of  force  are  such  that 

—  9*  —  i  then  the  work  done  on  a  particle  moving  between  two  points 
cy        dx 

depends  upon  the  path  of  the  particle,  the  work  done  on  a  particle  moving 

around  a  closed  path  is  not  zero,  and  there  exists  no  force  function. 

Such  a  force  is  called  a  nonconservative  force. 

Ex.  7.  Heat.  If  a  substance  is  brought,  by  a  series  of  changes  of  tem- 
perature, pressure,  and  volume,  from  an  initial  condition  back  to  the  same 
condition,  the  amount  of  heat  acquired  or  lost  by  the  substance  is  the 
mechanical  equivalentof  the  work  done,  and  is  not  in  general  zero.  Hence 
the   Line  integral  Q  =    C(M<lt  +  X</r)  around  a  (dosed  curve  is  not   zero. 

and  there  exists  no  function  whose  partial  derivatives  are  .1/  and  N.  In 
fact,  the  heat  Q  is  not  a  function  of  /  and  v,  not  being  determined  when 
t  and  v  are  given. 

172.  Differentiation  of  composite  functions.  It  is  frequently 
necessary  to  differentiate  with  respect  fco  a  variable  a  Function 
of  a  function  of  that  variable.  Several  cases  of  this  will  now 
be  discussed. 

1.  Consider /(h),  where  u  =  $(x). 

Then  #«#*/*(«)*!.  (1) 

dx      dudx     J   K  Jdx  K  J 

This  has  been  proved  in  §  82. 

2.  Consider /(m),  where  ?/  =  (/>(>,  //). 

T>1  Of         df  C II  .,  'II  .0. 

Then  ti  =  t,^  =  >(",,r-  <2> 

The  proof  of  this  formula  is  like  that  of  (1),  the  only 
difference  being  that 

T  .     Af     cf        T  .     Am      du 
Lun-^-=f-,      Lim—  =  —  • 

Ax      ox  Ax      ex 


358 

PARTIAL  DIFFERENTIATION 

Ex.    1.     If  2  = 

sin-, 

y 

dx 

Place  u  =  -  • 

y 

Then 

by  (2), 

dz            x  d    /x\      1        x 
—  =  cos 1  - )  =  -  cos  -  • 

dx             y  dx  \y/       y        y 

Similarly, 

dz             x  e    (x\            x         x 
—  =  cos 1  -  1  = cos  -  • 

dv         y < y  w        y       y 

Ex.  2.    If  z  =  f(x"  +  y'2),  show  that  x  —  -  y—  =  0. 

dy       ■   ax 

Place  u  =  x-  +  y'2.    Then 

f=f(u)^  =  2xf(u)> 

ex  dx 

dz  du 

—  =f(u)  —  =  2yf(u); 

dy      J    V    }dy  JJ   V    h 

whence  x  — '  —  y  —  =  0. 

By      J  dx 

3.   Consider /(w,  v),  where  ?£  =  0(V),  v  =  -\jr(x}. 
From  (6),  §  164,  with  a  change  of  letters, 

/}/*  ?*  f 

Af  =  —  An  +  —  Av  +  e,  Au  +  e,  Av. 
du  dv 

If  we  divide  by  Ax,  and  take  the  limit,  we  have 

df_cfdudfdv 
dx      du  dx      dv  dx 


Ex.  3.    Let  z  —  tan  - ,  where  u  =  ;r2,  r  =  loo;. 


I  roni  (2),  —  =     sec2  -    —    -     =  -  sec2  - , 

du       \         v}  du  \vi        v  o 

dz       I      .,  u\    8    ln\  u        .u 

—  =  ( sec2  -)  —  (-)= sec2  - 

dc       \         vj  do  \vl  v'"  v 

Hence,  from  (3),      —  =  -  (sec2  -)  2x  —  —  (sec2-)  - 
dx       v  \         vj  v'2  \         v/  x 

2  x  loo-  X  _  x  I    .,2    \ 

sec2 


(log  xf  \log  xl 


COMPOSITE  FUNCTIONS  359 

4.  Consider  _f(u,  v),  where  u  =  <j>(x,  y),  v  =  ty(x,  y~). 

Then  %JL*L  +  %.* 

cx      at  cx      cv  ex 

\  (4) 

cf      cf  cu      cf  cv 

by      cu  dy      dv  dy 
The  proof  is  like  that  of  (3). 

c -       d- 
Ex.  4.    If  z  =f(x  -  y,  y  -  x),  prove  —  +  __  =  0. 

Place  x  —  y  —  «,  y  —  x  =  v.    Then  z  =  /'(«,  v),  and 
cz  _  cf  du       if  dv  _  cf  _  cf 
cx      in  (i-      dv  dx      du      dv 
dz  _  df  du      dfdv  __  df      df 
£#       8u  dy        cr  c;i  cu        c  r 

By  addition  the  required  result  is  obtained. 

Ex.  5.    Let  it  be  required  to  change    !    and  -■  ■  from  rectangular  coor- 

dx  By 

dinates  (x,  //)  to  polar  coordinates  (r,  #),  where  ./•  =  rcosO,  y  —  /-sin#,  and 
/is  a  function  of  x  and  y. 

From  Ex.5.  §16 


whence 


cr 
dx 

= 

cos  9, 

cr 

dy 

=  sin  0, 

cO 

dx 

= 

sin 

e 

d0 
dy 

_  cos 
r 

df 

cr 

ci- 

CO8  0- 

df 
d$ 

sin8 

r 

df 

= 

cf 
cr 

>ii 

0  + 

df 

COS* 

r 

5.  If,  in  (4),  we  multiply  the  equations  (4)  by  dx  and  dy,  add 
the  results,  and  apply  the  definition  of  §  104,  we  have 

df^&du  +  tfdv. 

cu  cv 

This  result  is  easily  generalized.  Hence  the  form  of  the  differ- 
ential df  is  the  same  whether  the  variables  used  are  independent 
or  not. 

6.  Higher  derivatives  of  a  composite  function  may  be  found 
by  successive  applications  of  the  foregoing  formulas. 


360  PARTIAL  DIFFERENTIATION 

The  extension  of  all  the  foregoing  relations  to  cases  involving 
more  variables  is  obvious. 

02  y       32  y 
Ex.  6.    Required  to  express  — -  H in  polar  coordinates,  where  V  is 


a  function  of  x  and  y. 


dx2        dy2 


„         v      .                    dV      dV        ,.     aFsin0 
t  rom  JirX.  5,  —  =  —  cos  a -  , 

dx        dr  dd      r 

BV      dV    .     a  ,   dVcoad 
dy        dr  dd      r 

Then,  by  (4), 

d2V      dVdV        a      BVain'Oldr  ,    dVdV        Q      aFsin0~la0 
Vd2V        a       d2V  sin0  ,   a  F  sin  01        a 

-      ^V  cos  &  —  TT^Ti ^  ~K  7T    cos  " 

\_dr2  drdd     r  dd      r2   J 

,  ra2F       Q    e2Fsin0    dV  .  a    aF  cos  01/     sin0\ 

+    -k  cos  0 xr —  sin  0 x 

LdrdO  dd2      r  or  dd      r    ]\  r    J 

&V       9A     0  a2  F  sin  0  cos  0      ?2  Fsin20  ,  dV  sin20     o5Fsin0cosi 

=  ^co^-2^0 — r—  +  w—  +  -d7-T-  +  2Te^>— 

Similarly, 
d2V      d2V  .   2/1  ,  n  c2Fsin0cos0  ,  a2Fcos20  ,  aFcos20      oaFsin0cosi 
dy*        dr2  drcd         r  dd*      r2  dr       r  dd         r2 


drcd 

r 

d2V 
dx2 

a2  f  _  i 

dy* 

d2v  ,  l  a2F  ,  l  aF 

Hence  — -  H =  — -  +  —  — r-  H 

3/-2       r2  802       r  a/- 

Ex.  7.     If  z  =  f,  (x  +  at)  +  L  (x  -  at),  show  that  —a  =  a2  — . 

T      i.  TU  3,i  1        S"  Sl'  1        ^W 

Let  x  +  at  =  u,  x  —  at  =  v.    1  hen  —  =  1,  —  =  a,  —  =  1,  —  =  —  a, 

dx  dt  dx  dt 

,  ,    /ox  a*     ay:  au  ,  rf/2  to     .//;  ,  a/8 

and,  by  (2),  —  =  41 [- _Zl  _  =  ^1  +  ^2 , 

ax      «u  3x      aw  ox      a«       do 

Bz_dfk  a_«     ^to_  a(]f1_a(]f1 

dt       du    dt        dv    dt  du  do 

Differentiating  these  equations  a  second  time,  we  have 

d2z  =  d2fx  du       d%  do  =  d%    (    d%  ^ 
dx2       du2  dx       dv2   dx       du2        do2 

a<2       "  tf«2    a<       "  rfw2    dt       U    du2       °    dv2  ' 
By  inspection  the  required  result  is  obtained. 


PROBLEMS  361 

PROBLEMS 


Find  ;r-  and  —  >  when 
1.  .-- ^- 


4.   s  =  log  (>/  +  Vy2  - 

-a?). 

X  u 

5.  s  =  sin  — ! —  • 
x  —  y 

6.  «  =  e"i  +  log|- 

Va^  +  y2 

2.  s  =  tan-1-- 

x 

.x  —  y 

3.  «==  sin"1 z« 

a; 

7.  If  z  =  sin  (a-2  —  2 a*?/  +  ?/),  prove  -r-  +  t^1  =  0. 

V  9    x    9  »  *  OX         ClJ 

8.  From  2  a-2  -  3  if  +  G  .77/  +  2  .~2  =  0,  prove  a-  r-  +  y  j-  =  ». 

-    •     V  &  0«       « 

9.  If  z  =  ex  sin  -  >  prove  a-  ^-  +  y  ^-  =  0. 

x    L  ex      9  dy 

10.  If  z  =  y2  +  tan  Oj),  prove  x2  ~  + y -^=  2  y2. 

11.  If  s2  =  log  (a-?/)  +  sin-1  -  >  prove  sa;  ^  +  ~y  5-  =  1. 

0  v  *'  x  ex         '    ry 

12.  If  z  =  -5 =  e(l2  +  ^,  prove  y  ^  -  x  f1  =  0. 

x1  +  //'  0a-  dy 

13.  If  z  =  V?/2  —  a;2  sin-1  -  >  prove  jb  ^-  +  y  s~  —  «■ 

14.  Show  that  the  sum  of  the  partial  derivatives  of 

u  =  (x  —  y)  (y  —  z)(z  —  x)  is  zero. 

15.  Ifa-  =  «  +  r,y  =  ;-;,find(^and(|); 

16.  If  x  =  er  sin  0,  y  =  er  cos  $,  find  (  —  )  and  f  - -)  • 

17.  If  x  =  e"  sec  r,  y  —  eu  tan  v,  find  (5-)  and  (  — )  • 

18.  If  a;  =  re",  y  =  re'6,  find  f  ^-J  and  l-^J  • 

19.  If  «  =  (a-2  +  if)  tan-1  ^,  find  « 


362  PARTIAL  DIFFERENTIATION 

l  :;.v 

20.  Ii'  z  =  ey  sin  (x  —  ?/),  find 

21.  If  z  =  log(xa  +  y2),  prove  ^4  +  Tl  =  0- 

22.  If  ;■  =  tan  (y  4-  aa)  +  (//  —  a#)^j  prove  ~  =  a?  — ~2- 
Verify  ^— r-  =  ;— tt- 1  when  : 


23.  »  =  a^  +  2ye*. 

24. 


"^ 


25.  g  =  log  (a  +  Vy2  +  x*). 

V./'2  +  v2 

26.  g  = =!-^-- 


+  2/ 

27.  Calculate  the  numerical  difference  between  Ag  and  rfg,  when 
g  =  cb8  +  y3  -  3  afy,  a;  =  2,  y  =  3,  A*  =  cZz  =  .01,  and  Ay  =  </y  =  .001. 

28.  The  hypotenuse  and  one  side  of  a  right  triangle  are  respec- 
tively 5  in.  and  4  in.  If  the  hypotenuse  is  decreased  by  .01  in.  and 
the  given  side  is  increased  by  .01  in.,  find  the  total  change  made  in 
the  third  side,  the  triangle  being  kept  a  right  triangle.  Find  the 
error  that  would  be  made  if  the  differential  of  the  third  side  corre- 
sponding to  the  above  increments  were  taken  for  the  change. 

29.  A  right  circular  cylinder  has  an  altitude  10  ft.  and  a  radius 
5  ft.  Calculate  the  change  in  its  volume  caused  by  increasing  the 
altitude  by  .1  ft.  and  the  radius  by  .01  ft.  Calculate  also  the  differ- 
ential of  volume  corresponding  to  the  same  increments. 

30.  A  triangle  has  two  of  its  sides  8  in.  and  10  in.  respectively,  and 
the  included  angle  is  30°.  Calculate  the  change  in  the  area  caused 
by  increasing  the  length  of  each  of  the  given  sides  by  .01  in.  and 
the  included  angle  by  1°.  Calculate  also  the  differential  of  area 
corresponding  to  the  same  increments. 

31.  The  distance  between  two  points  A  and  B  on  opposite  sides 
of  a  pond  is  determined  by  taking  a  third  point  C  and  measuring 
AC  =  80  ft.,  BC  =  100  ft.,  and  BCA  =  60°.  Find  the  greatest  error 
in  the  length  of  AB  caused  by  possible  errors  of  6  in.  in  both  AC 
and  BC,  assuming  that  powers  of  the  errors  of  measurement  higher 
than  the  first  may  be  neglected. 

32.  The  distance  of  an  inaccessible  object  A  from  a  point  B 
is   found  by   measuring  a  base  line  BC  =  100  ft.  and  the  angles 


PROBLEMS  363 

CBA  =  a  =  30°  and  EC  A  =  /3  =  45°.  Find  the  largest  possible  error 
in  the  length  of  AB  caused  by  errors  of  1"  in  measuring  a  and  (3, 
assuming  that  powers  of  the  errors  of  measurement  higher  than  the 
first  may  be  neglected. 

33.  The    density   D  of   a   body   is   determined   by  the   formula 

D  = -.)  where  w  is  the  weight  of  the  body  in  air  and  w'  the 

w  —  w'  b  J 

weight  in  water.    If  w  =  243,600  gr.  and  w'  =  218,400  gr.,  what  is 

the  largest  possible  error  in  D  caused  by  an  error  of  o  gr.  in  w  and 

an  error  of  8  gr.  in  w',  assuming  that  powers  of  the  errors  of  w  and 

w'  higher  than  the  first  may  be  neglected  '.' 

34.  If  the  electric  potential  !'  at  any  point  of  a  plane  is  given  by 
the  formula  V=  log  Vx'2  +  >/2,  find  the  rate  of  change  of  potential  at 
any  point:  (1)  in  a  direction  toward  the  origin  ;  (2)  in  a  direction  at 
right  angles  to  the  direction  toward  the  origin. 

35.  If  the  electric  potential   I' at  any  point  of  the  plane  is  given 

by  the  formula  V=  log —  >  find'the  rate  of  change  of 

BV(x  +  a)"  +  y* 

potential  at  the  point  (0,  a)  in  the  direction  of  the  axis  of  //.  and  at 

the  point  (a,  a)  in  the  direction  toward  the  point  (—  a,  0). 

36.  On  the  surface  «  =  2tan-1— ,  find  the  slope  of  the  curve 
through  the  point  (1,  1.  t  )  whose  plane  makes  an  angle  of  30°  with 
the  plane  XOZ. 

37.  On  the  paraboloid  z  =  2  .r1  -p  :>  //-,  what  plane  section  perpen- 
dicular to  the  plane  XOY  &xid  through  the  point  (2,  1,  11)  will  cut 

out  a  curve  with  the  slope  zero  al   that   point'.' 

38.  In  what  direction  from  the  point  (.'•,.  //,)  is  the  directional 
derivative  of  the  function  z  =  hey  a  maximum,  and  what  is  the 
value  of  that  maximum  derivative? 

39.  Find  a  general  expression  for  the  directional  derivative  of 
the  function  u  =  e~ ■"  sin  x  +  -  e~ 3 v  sin  3  x  at  the  point  ( —  j  0  ).  Find 
also  the  maximum  value  of  the  directional  derivative. 

40.  If  s  =  4  .r2  +  2  //"  and  y  =  -,  find  ^- 


43. 


364  PARTIAL  DIFFERENTIATION 

v 

41.  If  a  point   moves  on  the  surface  *  =  k  tan-1  -  so  that  its 

X  </-■ 

projection  on  the  XOY  plane  is  the  circle  x2  +  if  =  a2,  find  —  • 

42.  Find  -y-  from  the  equation  x  =  cex. 

Find  -j-  from  the  equation  2  log  a;  +  (sin-1-]  =  c. 

d?         dz 

44.  Find  y  and  y  '  when  (a  +  //)  (y  +  «)  (,~  +  sc)  =  c. 

45.  Find  ^-  and  tt1  >  when  a"3  +  ya  +  ,-s2  —  log  v/,?2  =  c. 

ex  OlJ 

d-  dz 

46.  Find  y  and  Y'  when  ^3  +  ^  +  y^z  +  ^  =  °" 

47.  Find  ^  and  y ,  when  (x2  +  y2  +  *2)3  =  27  xyz. 

48.  Find  the  equations  of  the  tangent  plane  and  the  normal  line 
to  the  ellipsoid  x2  +  3  y2  +  2  z2  =  9  at  the  point  (2,  1,  1). 

49.  Find  the  equations  of  the  tangent  plane  and  the  normal  line 
to  the  surface  xy  4-  yz  +  «*  =1  at  the  point  (1,  0,  1). 

50.  Find  the  equations  of  the  tangent  plane  and  the  normal  line 
to  the  surface  z  =  (ax  +  by)'2  at  the  point  (xv  yv  zx). 

51.  Find  the  tangent  plane  to  the  cone  x2  +  y2  —  z2  =  0  and  prove 
that  it  passes  through  the  vertex  and  contains  an  element  of  the  cone. 

52.  Show  that  the  sum  of  the  squares  of  the  intercepts  on  the 
coordinate  axes  of  any  tangent  plane  to  the  surface  x*  +  y*  4-  z1  =  a5 
is  constant. 

53.  Show  that  any  tangent  plane  to  the  surface  z  =  Jcxy  cuts  the 
surface  in  two  straight  lines. 

54.  Find  the  equations  of  the  tangent  line  and  the  normal  plane 
to  the  curve  xyz  =  1,  y'2  =  x  at  the  point  (1,  1,  1). 

55.  Find  the  equations  of  the  tangent  line  and  the  normal  plane 
to  the  curve  x  =  sin  z,  y  =  cos  z  at  the  point  (l,  0,  —)• 

56.  Find  the  equations  of  the  tangent  line  and  the  normal  plane 
to  the  curve  of  intersection  of  the  cylinders  x2  +  y2  =  25,  v/2+s2  =  25 
at  the  point  (4,  3,  —  4). 


PROBLEMS  365 

57.  Find  the  equations  of  the  tangent  line  and  the  normal  plane 
to  the  curve  of  intersection  of  the  ellipsoid  24.r2  +  16//--|-3.?2  =  288 
and  the  plane  2a;  +  8y-f5«  =  0atthe  point  (2,  —  3,  4). 

?/ 

58.  Find  the  angle  at  which  the  helix  x2  +  if  =  a2,  z  =  k  tan-1'- 

intersects  the  sphere  x2  +  if  +  z2  =  r(r  >  a). 

59.  Find  the  angle  at  which  the  curve  f  —  z2  =  a2,  x  =  l>(y  +  z) 
intersects  the  surface  x2  +  2  zy  —  c2. 

60.  Find  the  minimum  value  of  the  function  z  =  £x2  —  3xy 
+  9y2  +  5x  +  15y  +  16. 

61.  An  open  rectangular  cistern  is  to  be  constructed  to  hold 
1000  cu.  ft.  Required  the  dimensions  that  the  cost  of  lining  should 
be  a  mininium. 

62.  Divide  the  number  a  into  three  parts  such  that  their  product 
shall  be  the  greatest  possible. 

63.  Find  a  point  in  a  plane  quadrilateral  such  that  the  sum  of  the 
squares  of  its  distances  from  the  four  vertices  is  a  minimum. 

64.  Find  the  volume  of  the  greatest  rectangular  parallelepiped 
inscribed  in  an  ellipsoid. 

65.  Find  by  calculus  the  point  in  the  plane  2  ./•  -f  ■">  //  —  ('. .-.  -(-5  =  0 
which  is  nearest  the  origin. 

66.  Find  the  points  on  the  surface  2rJ  4-  1  f  —  .-;'-  —  (/>./■  +  5 y 
-(- 18  =  0  which  art'  nearesl  the  origin. 

67.  Find  the  highest  point  on  the  curve  of  intersection  of  the 
hyperboloid  aj9  +  //"  —*-'"  =  1  and  the  plane  as  +  y  +  2«  =  0. 

68.  Find  the  volume  of  the  greatest  rectangular  parallelepiped 
which  can  be  inscribed  in  a  right  elliptic  cone  with  altitude  h 
and  semiaxes  of  the  base  a  and  b,  assuming  that  two  edges  of  the 
parallelepiped  are  parallel  to  the  axes  of  the  base  of  the  cone. 

69.  Through  a  given  point  (1,  1,  2)  a  plane  is  passed  which  with 
the  coordinate  planes  forms  a  tetrahedron  of  minimum  volume.  Find 
the  equation  of  the  plane. 

70.  Find  the  point  inside  a  plane  triangle  from  which  the  sum  of 
the  squares  of  the  perpendiculars  to  the  three  sides  is  a  minimum. 
(Express  the  answer  in  terms  of  K,  the  area  of  the  triangle ;  a,  b,  c, 
the  lengths  of  the  three  sides  ;  and  x,  y,  z,  the  three  perpendiculars 
on  the  sides.) 


366  PARTIAL  DIFFERENTIATION 

Prove   that  the  following  differentials  are  exact  and  find  their 
integrals : 

7 1 .  (5  x*  -  3  afy  +  2  x,f)  dx  +  ( 2  xh,  -  s8  +  5  ,f)  dy. 

72.  (y  +  -\dx+(x  +  -\dy. 

73  al  +  2?  +  aV  2^±1 

ay  «y     * 

,r2    +    y3  Q  J'3     , 

75.   !i,fe  _/£*-? +  ^W 

y         \y"        y/  l 

77.  (cos22C  —  //  sin  x)  da;  +  cos  a-  dy. 

78.  e2  sin (x  +  y) da;  +  e'1  [sin  (a;  +  y)  —  y  cos  (a:  +  y)~\dy. 

(0,0) 

(1)  along  a  straight  line, 

(2)  along  a  parabola  with  its  axis  on  OX. 

Jr«a.o) 
[(.r2  +  y"2)^.r  +  xdy\ 
(0,1) 

(1)  along  a  straight  line, 

(2)  along  a  circle  with  its  center  at  0. 

r&,  i) 

81.  Find  the  value  of  j        [y*dx  +  (xy  +  y")dy], 

«7(0,  0) 

(1)  along  a  parabola  with  its  axis  on  O.Y, 

(2)  along  a  broken  line  consisting  of  a  portion  of  the  ,r-axis 

and  a  perpendicular  to  it. 

82.  Find  the  value  of    /         I     ,"      "     „  +    ?' 


f'7  xdx 


r; 

(1)  along  the  curve  x  =  t,  y  =  t2, 

(2)  along  a  broken  line  consisting  of  a  portion  of  the  a--axis 

and  a  perpendicular  to  it. 


83.  Find  the  value  of 


J/">(5, 
(3,0) 


PROBLEMS  30' 

(5,  4)  _  y  (JX  _|_  x  dy 


Vx2  -  f 

(1)  along  the  curve  x  =  3  sec  0,  y  =  3  tan  0, 

(2)  along  a  broken  line  consisting  of  a  portion  of  the  cc-axis 

and  a  perpendicular  to  it. 

84.  Find,  by  the  method  of  Ex.  3,  §  171,  the  area  of  the  four- 
cusped  hypocycloid  x  =  a  cos8<f>,  y  =  a  sin8<£. 

85.  Find,  by  the  method  of  Ex.  3,  §  171,  the  area  between  one 

(arch  of  a  hypocycloid  (§  58)  and  the  fixed  circle. 
86.  Find,  by  the  method  of  "Ex.  .">,  ^  1 7 1 .  the  area  between  one 
arch  of  an  epicycloid  (§  57)  and  the  fixed  circle. 

87.   If  u  =f(x,  y)  and  y  =  F(x),  find 


dx* 
dx*\t  // '         dxi  1 1  dxdy     < '/    i  x 

W' 


T„    ,  „  '/"'/         cx~  \dy/        dxdy  dxdy     dv 

88.  If  f(x,  y)  =  0,  prove  —7'  - 

[cy 

/x\  t -  d~ 

89.  If  z  =/(-),  prove  x    '  +yi~  =  0. 

90.  If  f(lx  -f  my  +  nz,  x2  -+-  y'1  +  »9)=  <>,  prove 

I  ly  -  mx)  +  (ny  -  mz)  ^  +  (A-  -  nx  |  ^  =  0. 

91.If«  =  ^  +  2/(i  +  logy),Prove|  =  2y-^|. 

?/\  ,  ,M  -id'z  ,  o„..  ^*    ,    ■•''-'■- 


93.  If  c  =  x<f>  K   +  ^r  •'    ,  ] >rove  x2 -T-T,  +  2  xy  -=-=r  +  f  —  =  0. 
\xj      r\x/  <j-  dxdy      ■    ( y- 

94.  If  z  =  <£(.r  -+-  iy)  -\-  \p(x  —  iy),  where  i  =  V  —  1,  prove 

d-z      c2z      A 
— ;  +  —  =  0. 


95.  If  V  is  a  function  of  ;•  only,  where  r  =  Vsc3  +  y2,  find  the  value 
of  ^-g  +  ~-^  m  terms  of  r  and  T . 


3G8  PARTIAL  DIFFERENTIATION 

96.  If  x  =  eu,  y  —  ev,  and  z  is  any  function  of  x  and  y,  find  the 

d2z  d2z  dz  dz 

value  of  x2  TT-r,  +  y2  tth  +  x  ^ — h  V^~  in  terms  of  the  derivatives  of  z 
ox"  dy  dx  dy 

with  respect  to  w  and  v. 

97.  If  x  =  u  +  w,  <y  = ->  and  F  is  any  function  of  x  and  ?/, 

0a  V      d*V        2d2V 
prove  aa  -^-5  —  -5-5  =  a-  5— «-  • 

98.  If  .x  =  e"  cos  v,  y  =  e"  sin  ?;,  and  F  is  any  function  of  x  and  y, 

d2V 
find  r — —  in  terms  of  the  derivatives  of  V  with  respect  to  x  and  y. 
dado 

99.  If  x  =  eu  cos  v,  y  =  e"  sin  -v,  and  F  is  any  function  of  x  and  y, 
d2V      d2V  0./PV  ,   d2F\ 

Prove^  +  ^  =  e~"'W  +  W 

e«  _i_  g-o  ^0 e-e 

100.  If  x  =  1 — — >  y  =  ) n~ — >  and  F  is  any  function  of 

d2V      fflV      d2V       1  d2V      ldV 

x  and  y,  prove  ■=-=  —  -z-=  =  -^-r g  ^  i ^—  • 

•n  l  dx1        dy1        Ci-2        r2  d62       r  dr 

101.  If  x  =  ev  sec  u,  y  =  ev  tan  it,  and  <£  is  any  function  of  x  and  y, 

/  d24>       d<f>\  (d24>      d2<f>\      I  2        a  a2<*> 

102.  If  x  -t  y  =  2  ee  cos  <j>,  x  —  y  =  2  ie8  sin  <£,  and  F  is  any  func- 

'         a2r    a2F     ,      a2F 

tion  of  x  and  y,  prove  _  +_  «  4ay ^ 

103.  If  x=f(u,  v)   and  y=<f>(u,  v)  are  two  functions   which 

satisfy  the  equations  ^-  =  -r21  >  ^-  =  —  -^->  and  V  is  any  function  of 

d2V      d2V      (c2V      82V\V/df\2     (d/Vl 
x  and  y,  prove  ^  +  -^  =  (^  +  -^)[{^)  +{Fu)  [ 


CHAPTER  XVI 


MULTIPLE  INTEGRALS 


173.   Double  integral  with  constant  limits.    By  definition, 
f(x)  dx  =  Lira  2)  /OO  Ax, 


and  f(x)  dx  is  the  element  of  the  integral.  In  the  problems  of 
Chapter  XIII  it  has  been  possible  to  form  the  element  f(x)dx 
immediately  by  elementary  theorems  of  geometry  and  mechanics. 
There  are  problems,  however,  in  which  it  is  advantageous  to 
determine  the  element  itself  as  a  definite  integral. 

For  example,  let  us  find  the  volume  bounded  by  the  planes 
z  =  0,  .r  =  a,  x  =  b(a  <b~),  y  =  c,  y  =  d(c  <  d),  and  the  surface 
z=f(x,y)  (fig.  212) 
which  lies  entirely 
on  the  positive  side 
of  the  XOY  plane 
for  the  volume  to 
be    considered. 

Divide  the  dis- 
tance b  —  a  on  OX 
into  n  equal  parts 
Arc;  thus  giving  x 
the  series  of  values 

«,     xx  =  a  +  A.r,     r, 2  =  xx  +  A.v, 

Through  the  points  thus  determined  on  O X  pass  planes  par- 
allel to  YOZ,  thus  dividing  the  required  volume  into  slices, 
such  as  LM. 

Divide  the  distance  d  —  c  on  OY  into  m  equal  parts  Ay,  thus 
giving  y  the  series  of  values 

c,     y1  =  c  +  Ay,     y2  =  yx  +  Ay, 
369 


370  MULTIPLE  INTEGRALS 

Through  the  points  thus  determined  on  OY  pass  planes  par- 
allel to  XOZ,  thus  subdividing  the  slices  into  volumes,  such  as 
NQ,  each  of  which  stands  on  a  base  NB,  Ax  At/  in  area. 

If  N  has  coordinates  (z„  yj),  NP  =f(xv  yj),  and  the  volume  of 
a  prism  with  NR  as  a  base  and  NP  as  altitude  \s,f(xv  y^AxAy. 

If  we  hold  x  equal  to  .r.,  and  give  y  the  values  c,  yv  y0,  •  •  • 
in  succession,  and  take  the  limit  of  the  sum  as  m  =  go,  we  have 


UmVf(xnyi)AxAy=  I    . 


f(xny)Ax<h,  (1) 

as  an  approximate  expression  for  the  volume  of  the  slice  LM. 

Using  the  definite  integral  (1)  as  an  element,  we  now  assign 
to  x  the  values  a,  xv  x2,  •  ■  •  in  succession  and  take  the  limit 
of  the  sum  as  n  =  cc.    The  result  is 

l^  X  jf /o*,  y)  *u  A.r  =£(f"f^  y)  dy)d*'     (2) 

which  is  the  required  volume. 

Removing  the  parentheses,  we  shall  write  (2)  in  the  form 

'lf(x,y)dxJy,  (3) 


If 


where  the  summation  is  made  in  the  order  of  the  differentials 
from  right  to  left,  i.e.  first  with  respect  to  y  and  then  with 
respect  to  x,  and  the  limits  are  in  the  same  order  as  the  differ- 
entials, i.e.  the  limits  of  y  are  c  and  d,  and  the  limits  of  x  are 
a  and  b* 

Referring  to  fig.  212,  we  see  that  we  could  have  made  the  sum- 
mation first  with  respect  to  x,  thereby  finding  an  approximate 

*  Still  another  form  of  writing  (2)  is  f    dx  J    f(x,  y)  dy,  in  which  the  order 

of  summation  is  first  with  respect  to  y  and  then  with  respect  to  x. 

I    /(/,  y)dydx,  which 

is  merely  (2)  with  the  parentheses  removed.  In  this  form  it  is  to  be  noted  that 
the  limits  and  the  differentials  are  in  inverse  orders,  and  that  the  order  of 
summation  is  the  order  of  the  differentials  from  left  to  right,  i.e.  first  with 
respect  to  y  and  then  with  respect  to  x.  In  this  text  this  last  form  of  writing 
the  double  integral  will  not  be  used.  In  other  books  the  context  will  indicate 
the  form  of  notation  which  the  writer  has  chosen. 


DOUBLE  INTEGRALS 


371 


expression  for  the  volume  of  a  slice  bounded  by  two  planes  par- 
allel to  the  YOZ  plane.  The  final  summation  would  have  been 
with  respect  to  y,  the  result  being  the  volume  expressed  by  (3). 
If  this  order  had  been  followed,  the  result  would  have  appeared 
in  the  form 

jf  (jf/o*  ^ch)(i'/=f£/{-"  ■>n,j'/,ir-    (4) 

Integrals  (3)  and  (4)  are  called  double  definite  integrals,  t la- 
limits  in  this  case  being  constants.  As  any  function  f(x,  y ) 
may  be  represented  graphically  by  the  surface  2  =/(./■,  //),  we 
are  led  to  the  general  definition  of  the  double  definite  integral, 

J  j  /('•,  y)dydx  = C  J ./(,-.  y)dxdy, 

as  equal  to  the  limit,  as  m  and  n  are  both  increased  indefi- 
nitely, of  the  double  sum 


X    '  %/(&,  //,)A.rA//, 


(5) 


where  Ar,  A//,  and  (xc  //, )  have  the  meanings  already  defined. 
The  integral  is  called  the  double  integral  of  f(x,  // )  over  the 
area  bounded  by  the  lines  x  =  a,  x  =  b,  y  =  <;  y  =  d.  This 
definition  is  independent  of  the  graphical  interpretation,  and 
therefore  any  problem  which  leads  to  the  limit  of  a  sum  (5) 
involves  a  double  integral. 

174.  Double  integral  with  variable 
limits.  We  may  now  extend  the  idea 
of  a  double  integral  as  follows:  In- 
stead of  taking  the  integral  over  a 
rectangle,  as  in  §  173,  we  may  take 
it  over  an  area  bounded  by  any 
closed  curve  (fig.  213)  such  that  a 
straight  line  parallel  to  either  OX  or 
OT  intersects  it  in  not  more  than 
two   points.     Drawing  straight    lines 

parallel  to  OY  and  straight  lines  parallel  to  OX,  we  form 
rectangles  of  area  AxAy,  some   of  which  are   entirely  within 


372  MULTIPLE  INTEGRALS 

the  area  bounded  by  the  curve  and  others  of  which  are  only 
partly  within  that  area.     Then 

%%  f&  y-)  Ax  Ay,  (1) 

where  the  summation  includes  all  the  rectangles  which  are  wholly 
or  partly  within  the  curve,  represents  approximately  the  volume 
bounded  by  the  plane  XOY,  the  surface  z  =f(x,  y),  and  the  cyl- 
inder standing  on  the  curve  as  a  base,  since  it  is  the  sum  of  the 
volumes  of  prisms,  as  in  §  173.  Now,  letting  the  number  of  these 
prisms  increase  indefinitely,  while  Az=0  and  Ay  =  0,  it  is  evident 
that  (1)  approaches  a  definite  limit,  the  volume  described  above. 
If  we  sum  up  first  with  respect  to  y,  we  add  together  terms 
of  (1)  corresponding  to  a  fixed  value  of  x,  such  as  xt.  Then 
if  MB  is  the  line  x  =  :r.,  the  result  is  a  sum  corresponding 
to  the  strip  ABCD,  and  the  limits  of  y  for  this  strip  are  the 
values  of  y  corresponding  to  x  =  xi  in  the  equation  of  the  curve. 
That  is,  if  for  x  =  x{,  the  two  values  of  y  are  MA=f1(xis)  and 
MB  =/2(xt.),  the  limits  of  y  are  fx(x^)  and  f%(x^).  As  different 
integral  values  are  given  to  i,  we  have  a  series  of  terms  corre- 
sponding to  strips  of  the  type  ABCD,  which,  when  the  final 
summation  is  made  with  respect  to  x,  must  cover  the  area 
bounded  by  the  curve.  Hence,  if  the  least  value  of  x  for  the 
curve  is  the  constant  a  and  the  greatest  value  is  the  constant 
b,  the  limit  of  (1)  appears  in  the  form 

*Xf(x,y)dxdy,  (2) 

7i  (x) 

where  the  subscript  i  is  no  longer  needed. 

On  the  other  hand,  if  the  first  summation  is  made  with 
respect  to  x,  the  result  is  a  series  of  terms  each  of  which  corre- 
sponds to  a  strip  of  the  type  A'B'C'D',  and  the  limits  of  x  are 
of  the  form  ^(y)  and  </>„(j/)>  found  by  solving  the  equation  of 
the  curve  for  x  in  terms  of  y.  Finally,  if  the  least  value  of  y 
for  the  curve  is  the  constant  c  and  the  greatest  value  is  the 
constant  d,  the  limit  of  (1)  appears  in  the  form 

n«2(Z/) 
f(x,  y)dydx.  (3) 

.At/) 


J  a    J,\(. 


DOUBLE  INTEGRALS 


373 


While  the  limits  of  integration  in  (2)  and  (3)  are  different, 
it  is  evident  from  the  graphical  representation  that  the  integrals 
are  equivalent. 

So  far  in  this  chapter  f(x,  y)  has  been  assumed  positive  for 
all  the  values  of  x  and  y  considered,  i.e.  the  surface  z  =f(x,  y) 
was  entirely  on  the  positive  side  of  the  plane  XOY.  If,  however, 
f(x,  y)  is  negative  for  all  the  values  of  x  and  y  considered,  the 
reasoning  is  exactly  as  in  the  first  case,  but  the  value  of  the 
integral  is  negative.  Finally,  if  f(x,  y)  is  sometimes  positive  and 
sometimes  negative,  the  result  is  an  algebraic  sum,  as  in  §  81. 

175.  Computation  of  a  double  integral.  The  method  of  comput- 
ing a  double  integral  is  evident  from  the  meaning  of  the  notation. 

Ex.  1.    Find  the  value  of  f      I    xydxdy. 

Jo  Jo 

As  this  integral  is  written,  it  is  equivalent  to  j  (  j  xydyjdx,  the  integral 
in  parentheses  being  computed  first,  on  the  hypothesis  that  y  alone  varies. 

Ex.  2.    Find  the  value  of  the  integral  j  j  xydxdy  over  the  first  quadrant 

of  the  circle  x-  +  y2  —  a2. 

If  we  sum  up  first  with  respect  to  y,  we  find  ;i 
series  of  terms  corresponding  to  strips  of  the  type 
ABCD  (fig.  214),  and  the  limits  of  y  are  the  ordinate's 
of  the  points  like  A  and  B.  The  ordinate  of  .1  is 
evidently  0,  and  from  the  equation  of  the  circle  the 
ordinate  of  B  is  Va*  —  x'2,  where  OA  =  x.  Finally,  to 
cover  the  quadrant  of  the  circle  the  limits  of  x  are 
0  and  a.    Hence  the  required  integral  is 


i 


.i  i) 

Fig.  214 


Jo  Jo  J o  |_  2   Jo 

~2L  2         4 Jo 


374  MULTIPLE  INTEGRALS 

176.  Double  integral  in  polar  coordinates.  Let  us  assume  that 
we  have  a  function  f(r,  0)  expressed  in  polar  coordinates  and 
an  area  bounded  by  a  curve  (rig.  215)  which  is  also  expressed  in 
polar  coordinates.  As  in  §  1-11,  we  may  graphically  express  the 
function  by  placing  z  =f(r,  6),  where  the  values  of  z  correspond- 
ing to  assigned  values  of  r  and  6  are  laid  off  on  perpendiculars 
to  the  plane  of  r  and  6  at  the  points  determined  by  the  given 
values  of  r  and  6.  It  follows  that  the  graphical  representation 
of  f(r,  0)  is  a  surface. 

Let  us  now  try  to  find  the  volume  bounded  by  this  surface, 
the  plane  of  r  and  6,  and  the  cylinder  standing  on  the  curve 
as  a  base.  Proceeding  in  a  manner 
analogous  to  that  in  §  174,  we  divide 
the  area  into  elements,  such  as  ABCD 
(fig.  215),  by  drawing  radius  vectors 
at  distances  Ad  apart,  and  concentric 
circles  the  radii  of  which  increase  by 
Ar.  The  area  of  ABCD  is  the  dif- 
ference of  the  areas  of  the  sectors 
OBC  and  OAD.  Hence,  if  OA  =  r, 
area  ABCD=\  (r4-Ar)2A0-ir2A6>  =  (r-fe)  Ar  A0,  where  e=J-Ar. 

Then  the  volume  of  the  element  standing  on  the  base  ABCD  is 

f(r,  0)rArA<9+/(r,  0)eArA0, 
and  the  total  volume  is  the  limit  of  the  double  sum 

5J5jr/<>,  0)rArA0+/(r,  0)eArA0], 
or,  what  is  the  same  thing,  the  limit  of  the  double  sum 

X  X  f(r>  ^ r  Ar  Ae  =  fff(r>  e^> r  dr  de-        0-) 

If  the  summation  in  (1)  is  made  first  with  respect  to  r,  the 
result  is  a  series  of  terms  corresponding  to  strips  such  as  A1Bl  C\Diy 
and  the  limits  of  r  are  functions  of  6  found  from  the  equation 
of  the  boundary  curve.  The  summation  with  respect  to  6  will 
then  add  all  these  terms,  and  the  limits  of  6  taken  so  as  to 
cover  the  entire  area  will  be  constants,  i.e.  the  least  and  the 
greatest  value  of  6  on  the  boundary  curve. 


DOUBLE  INTEGKALS  375 

If,  on  the  other  hand,  the  summation  is  made  first  with  respect 
to  0,  the  result  is  a  series  of  terms  corresponding  to  strips  such 
as  A2B2C\2D2,  and  the  limits  of  6  are  functions  of  r  found  from 
the  equation  of  the  boundary  curve.  The  summation  with  respect 
to  r  will  then  add  all  these  terms,  and  the  limits  of  r  will  be  the 
least  and  the  greatest  value  of  r  on  the  boundary  curve. 

Now  f(r,  6)  may  be  any  function,  and  (1),  which  is  inde- 
pendent of  the  graphical  representation,  is  called  the  double 
definite  integral  over  the  area  considered.  Furthermore,  the  area 
of  ABCD  has  been  denoted  in  (1)  by  rdrd0,  i.e.  by  the  product 
of  AB  and  AD,  for  AB  =  dr  and  AD  =  rdO. 

Ex.    Find  the  integral  of  /•-  over  the  circle  r  =  2acos0. 

If  we  sum  up  first  with  respecl  to  r,  the  limits  are  0  and  '2  a  cos  6, 
found  from  the  equation  of  the  boundary  curve,  and  the  result  is  a 
series  of  terms  corresponding  to  sectors  of  the  type  A  OB  (fig.  216).  To 
sum  up  these  terms  so  as  to  cover  the  circle, 

the  limits  of  6  are and      •    The  result  is 


,£     "**-/.,,[3.    "* 

=  f\4:a*COB*dd$ 


Fig.  21G 


We  might  have  solved  this  problem  as  follows:  Since  the  Initial  line 
is  a  diameter  of  the  circle  and  the  values  of  r'1  at  corresponding  points  of 
the  two  semicircles  are  the  same,  it  is  evideni  that  the  required  integral 
is  twice  the  integral  taken  over  the  semicircle  in  the  first  quadrant. 

By  this  method  the  result  is 


J-.  ,,    jiiai 
o  Jo 


laif2co8i0dd 


Such  use  of  symmetry  as  was  made  in  the  second  solution  above  is  so 
often  of  advantage  that  the  student  should  always  note  when  there  is 
symmetry,  and  arrange  his  work  accordingly. 


376 


MULTIPLE  INTEGRALS 


177.  Area  bounded  by  a  plane  curve.  Let  us  in  (2),  §  174, 
denote  ,f\{x)  by  yx  and  f2(x)  by  y2,  and  omit  f(x,  y).  The 
result  is  ph  r„ 

I      I   'dxdy,  (1) 

J  a     J  V\ 

which  is  evidently  the  area  bounded  by  the  curve  in  fig.  213. 
But  nh  nyi  nb 

yjdx,  (2) 


where  (j/2  —  y^)dx  is  the  area  of  the  rectangle  ABCD. 
In  the  same  way  we  may  transform  (3)  of  §  174  into 


I 


(x^-x^dy, 


(3) 


which  will  represent  the  same  area  that  is  represented  by  (2), 
(.r2  —  x^)  dy  being  the  area  of  the  rectangle  A'B'  C'D'. 

It  is  evident  that,  if  the  area 
bounded  by  a  plane  curve  expressed 
in  rectangular  coordinates  is  found 
by  double  integration,  the  result  of 
the  first  integration  is  an  integral  of 
the  type  given  in  §  125. 

Ex.  Find  the  area  inclosed  by  the  curve 
(y- a; -3)2  =  4-^  (fig.  217). 

Solving  the  equation  of  the  curve  for  y 
in  terms  of  x,  we  have 


y 


x  +  3  ±  V4 


Accordingly  we  let  y1  =  x  +  3  —  V-i  —  x2 
and  y2  =  x  +  3  +  V4  —  x2,  whence  y„  —  y1  =  2  V4  —  x2,  and  take  for  the 
element  of  area  a  rectangle  such  as  ABCD.    Its  area  is  2  V4  —  x2dx. 

Since  the  curve  is  bounded  by  the  lines  x  =  —  2  and  x  =  2,  —  2  and  2 
are  the  limits  of  integration.    Hence  the  area  =   (    2  v4  —  x2dx  =  4  ir. 

In   like   manner  the   area  bounded  by   any  curve   in   polar 
coordinates  may  be  expressed  by  the  double  integral 


If' 


drd9, 


(4) 


MOMENT  OF  INERTIA  377 

the  element  of  area  being  that  bounded  by  two  radius  vectors 
the  angles  of  which  differ  by  A0,  and  by  the  arcs  of  two 
circles  the  radii  of  which  differ  by  A>\ 

We  may  transform  (4)  into  forms  similar  to  (2)  or  (3),  or 
we  may  make  the  double  integration,  substituting  both  sets  of 
lhnits  in  each  problem. 

If  the  first  integration  of  (4)  is  with  respect  to  r,  the  result 
before  the  substitution  of  the  limits  is  A  rdO,  which  is  exactly 
the  expression  used  in  computation  by  a  single  integration. 

178.  Moment  of  inertia  of  a  plane  area.  The  moment  of  inertia 
of  a  particle  about  an  axis  is  the  product  of  its  mass  and  the  square 
of  its  distance  from  the  axis.  The  moment  of  inertia  of  a  number 
of  particles  about  the  same  axis  is  the  sum  of  the  moments  of 
inertia  of  the  particles  about  that  axis.  From  this  definition  we 
may  derive  the  moment  of  inertia  of  a  lamina  of  uniform  thick- 
ness k,  and  of  density  p,  about  any  axis  as  follows  : 

Divide  the  surface  of  the  lamina  into  elements  of  area  dA. 
Then  the  mass  of  any  element  of  the  lamina  is  pkdA.  Let  Bi 
be  the  distance  of  any  point  of  the  /th  element  Prom  the  axis. 
We  may  then  take  as  the  moment  of  inertia  of  the  /th  element 
E'rpkdA,  the  exact  expression    evidently   being   (  R?  +  e^pkdA 

(§124).    If  the  lamina  is  divided  into  n  elements,  V  llfpk  dA 

is  an  approximate  expression  for  the  moment  of  inertia  of  the 
lamina.  Then,  if  /  represents  the  moment  of  inertia  of  the 
lamina,  i=n 

1=  Um  V  RfpkdA  =  jA'-p/cdA,  (1) 

"=x    1  =  1  J 

where  the  integration  is  to  include  the  whole  lamina. 

If  in  (1)  we  let  k=l  and  /o  =  l,  the  resulting  equation  is 


■/- 


dA,  (2) 


where  I  is  called  the  moment  of  inertia  of  the  plane  area  which  is 
covered  by  the  integration.  When  dA  in  (1)  or  (2)  is  replaced 
by  either  dxdy  or  rdrdO,  the  double  sign  of  integration  must 
be  used. 


378 


MULTIPLE  INTEGRALS 


Ex.  1.  Find  the  moment  of  inertia,  about  an  axis  perpendicular  to  the 
plane  at  the  origin.,  of  the  plane  area  (fig.  218)  bounded  by  the  parabola 
y-  =  -1  ((.c,  the  line  y  =  2  a,  and  the  axis  OY. 

We  divide  the  area  into  elements  by  straight  lines  parallel  to  OX  and  OY, 
Then  (1. 1  =  dx  dy,  and  R2  =  x2  +  //-,  whence  the  expression  for  the  moment  of 
inertia  of  any  element  is  (x2  +  y2)dxdy. 

If  the  integration  is  made  first  with  respect  to  x,  the  limits  of  that 

if 
integration  are  0  and  — ,  since  the  operation  is  the  summing  of  elements 

4  a 
of  moment  of  inertia  due  to  the  elementary  rectangles  in  any  strip  corre- 
sponding to  a  fixed  value  of  y ;  the  limit  0  is  found  from  the  axis  of  y, 

if 
and  the  limit  —  is  found  from  the  equation 
4  a 

of  the  parabola. 

Finally,  the  limits  of  y  must  be  taken  so 

as  to  include  all  the  strips  parallel  to  OX, 

and  hence  must  be  0  and  2  a. 


Therefore 


Jo     Jo 

Jo     \U)2    as     4     a      J 


178 
105' 


Y 

::::z 

/ 

/ 

u 

-L 

- 

f 

J 

0 

Fig.  218 

Ex.  2.  Find  the  moment  of  inertia  about  OTof  the  plane  area  bounded 
by  the  parabola  y'2  =  4  ax,  the  line  y  =  2  a,  and  the  axis  0  Y. 

Since  the  above  area  is  the  same  as  that  of  Ex.  1,  the  limits  of 
integration  will  be  the  same  as  there  determined,  but  the  integrand  will 
be  changed  in  that  x2  +  y2  is  replaced  by  x'2,  for  R  =  x. 


Hence 


fo°fo*«x*dydx 

m*C*dJf 

o 


Ex.  3.  Find  the  moment  of  inertia,  about  an  axis  perpendicular  to  the 
plane  at  O,  of  the  plane  area  (tig.  219)  bounded  by  one  loop  of  the  curve 
r  =  a  sin  2  6. 

We  shall  take  the  loop  in  the  first  quadrant,  since  the  moments  of 
inertia  of  all  the  loops  about  the  chosen  axis  are  the  same  by  symmetry. 


CENTER  OF  GRAVITY  379 

We  divide  the  area  into  elements  of  area  by  drawing  concentric  circles 
and  radius  vectors.  Then  dA  =  rdnlQ  (§  176),  and  R-  =  /■-,  whence  the 
element  of  moment  of  inertia  is  r^drdd. 

If  the  first  integration  is  made  with  respect  to  r,  the  result  is  the  moment 
of  inertia  of  a  strip  bounded  by  two  successive  radius  vectors  and  a  circular 
arc ;  hence  the  limits  for  r  are  0  and  a  sin  2  $.    Since  the  values  of  6  for 

the  loop  of  the  curve  vary  from  0  to  —  >  it  is 

evident  that  those  values  are  the  limits  for 
$  in  the  final  integration. 


Therefore       /  =    f  -    f"sl"    r*d9dr 
=  I  a*  C  *8in<  2  6<16 


Pig.  219 


179.  Center  of  gravity  of  plane  areas.  If  the  center  of  gravity 
of  any  physical  body  can  be  expressed  by  two  coordinates  x  and 
y,  we  proved  in  §  lo7  that 

/  xdm  I  >i dm 


P'"  ./' 


lm 


where  x  and  y  are  the  coordinates  of  the  point   at   which  the 
element    of   mass   <hn    may    l>r   regarded    as   concentrated. 
We  may  now  place 

d in  =  pil.nl '//,     or     dm  =  prdrd01 

where  p  is  the   mass   per  unit    area,   in    which    ease    the    above 
integrals  become  double  integrals. 

Ex.  1.    Find  the  center  oi  gravity  of  the  segment  »>t  the  ellipse  —  + 1-  =  1 

rt2       or 

cut  off  by  the  chord  through  the  positive  ends  of  the  axes  of  the  curve. 

This  is  Ex.  4,  §  lo7,  and  the  student  should  compare  the  two  solutions. 

The  equation  of  the  chord  is  bx  +  ay  =  ab. 

Dividing  the  area  into  elements  dxdy  (fig.  220),  we  have  dm  =  pdxdy, 
but  we  may  omit  p  since  it  is  constant.  Hence,  to  determine  x  and  y,  we 
.have  to  compute  the  two  integrals  CCxdxdy  and  CCydxdy  over  the  area 
A  CUD,  and  also  find  that  area. 


380 


MULTIPLE  INTEGRALS 


The  area  is  the  area  of  a  quadrant  of  the  ellipse  less  the  area  of  the 
triangle  formed  by  the  coordinate  axes  and  the  chord,  and  accordingly  is 

\(Trab)-  \ah=  \  ah {ir -2). 

For  the  integrals  the  limits  of 
integration  with   respect   to   y  are 


Vi  = 


ab  —  bx 


and 


b    /— 


yl  being  found  from  the  equation 
of  the  chord,  and  y2  being  found 
from  the  equation  of  the  ellipse. 
The  limits  for  x  are  evidently 
0    and    a. 


f"P' 

JO     J„b- 


x  dx 


dy=   fV^xVa2 

J  0       \fl 


f  fa     °  ydxdy  =  —    f°(-  b°-x"  +  dPz)dx 

Jo     Jab-bx  CpJo 


b2a. 


Therefore 


2  6 


3(*-2)  "      8(»-2) 

Ex.  2.    Find  the  center  of  gravity  of  the  area  bounded  by  the  two  circles 
r  =  a  cos  0,         r  =  b  cos  6.  (J>  >  a) 

It  is  evident  from  the  symmetry 
of  the  area  (fig.  221)  that  y  =  0. 

As  p  is  constant,  the  denominator 
of  x  is  the  difference  of  the  areas  of 
the  two  circles,  and  is  equal  to 


^ 


=  l  "(W-**). 


Since    x  —  r  cos  6,    and    the    ele- 
ment of  area  is  rdrdO,  the  numer- 


)r  of  x  be 

comes 

Fig 

221 

fl     fhC°Ser2 
"     77  J  a  cos  0 

cos  0d6dr  = 

K^3- 

a3)   f  2cos 

Odd 

= 

J»(8" 

-a3). 

Therefore 

x  = 

b2  +  ab  +  a2 

2  (b  +  a) 


AREA 


381 


180.  Area  of  any  surface.  Let  C  (fig.  222)  be  any  closed 
curve  on  the  surface  f(x,  y,  z)  =  0.  Let  its  projection  on  the 
plane  XOY  be  C.  We  shall  assume  that  the  given  surface  is 
such  that  the  perpendicular  to  the  plane  XOY  at  any  point 
within  the  curve  C  meets  the  surface  in  but  a  single  point. 

In  the  plane  XOY  draw  straight  lines  parallel  to  OX  and  OY, 
forming  rectangles  of  area  AxAy,  which  lie  wholly  or  partly  in 
the  area  bounded  by  C.  Through  these  lines  pass  planes  paral- 
lel to  OZ.  These  planes  will  intersect  the  surface  in  curves, 
which  intersect  in  points  the  z 
projections  of  which  on  the 
plane  XOY  are  the  vertices  of 
the  rectangles ;  for  example, 
M  is  the  projection  of  P.  At 
every  such  point  as  P  draw 
the  tangent  plane  to  the  sur- 
face. From  each  tangent  plane 
there  will  be  cut  a  parallelo- 
gram* by  the  planes 
drawn     parallel    to    <>Z. 

We  shall  now 
define  the  area 
of  the  surface 
/(•<■,  Ih  *0=0, 

bounded  by  the 

curve  C,  as  the  limit  of  the  sum  of  the  areas  of  these  parallelo- 
grams cut  from  the  tangent  planes,  as  their  number  is  made  ti> 
increase  indefinitely,  at  the  same  time  that  Ax==0  and  Ai/  =  0. 
It  may  be  proved  that  the  limit  is  independent  of  the  manner 
in  which  the  tangent  planes  are  drawn,  or  of  the  way  hi  which 
the  small  areas  are  made  to  approach  zero. 

If  AA  denotes  the  area  of  one  of  these  parallelograms  in  a 
tangent  plane,  and  7  denotes  the  angle  which  the  normal  to  the 
tangent  plane  makes  with  OZ,  then  (§  145) 

A.r  A^  =  AA  cos  7,  (1) 

*  This  parallelogram  is  not  drawn  in  the  figure,  since  it  coincides  so  nearly 
with  the  surface  element. 


8S2 


M  U  LTIPLE  IN  TEGlt ALS 


since  the  projection  of  A.'i   on  the  plane  XOY  is  A.rAv/.     The 
direction  cosines  of  the  normal  are,  by  (9),  §  168,  proportional 

dz   dz 


to 


dx    c// 


1 ;  hence 

cos  7 : 


1 


n^ar 


and  hence 


and 


a-'=^+(IJ+(SJa^       <2> 
x—xx^RSFM^    & 


According  to  the  definition,  to  find  A  we  must  take  the  limit 
of  (3)  as  A.i-=0  and  At//  =  0;  that  is, 


■114 


-'s)+@)** 


O) 


where  the  integration  must  be  extended  over  the  area  in  the 
plane  XOY  bounded  by  the  curve  C. 

Ex.  1.    Find  the  area  of  an  octant  of  a  sjihere  of  radius  a. 

If  the  center  of  the  sphere  is  taken  as  the  origin  of  coordinates  (fig.  223), 
the  equation  of  the  sphere  is 

x2  +  y2  +  z2  =  a2, 

and  the  projection  of  the  required 
area  on  the  plane  XOY  is  the  area 
in  the  first  quadrant  bounded  by 
the  circle 

y2  +  y2  =  a2 

and  the  axes  OX  and  OY. 

From  (1),      —=--, 
v  J       dx  z 

dy         z' 


^FWFW- 

Therefore    A  =         I  =  -ira  I    dx  =  - 

Jo  Jo  yVr  -./;--  u2       2       Jo  2 


AREA 


383 


Ex.  2.  The  center  of  a  sphere  of  radius  2  a  is  on  the  surface  of  a  right 
circular  cylinder  of  radius  a.  Find  the  area  of  the  part  of  the  cylinder 
intercepted  by  the  sphere. 

Let  the  equation  of  the 
sphere  be 

Xs  +  f  +  z*  =  1  a\   (1) 

the  center  being  at  the 
origin  (fig.  224),  and  let 
the  equation  of  the  cylin- 
der be 

f  +  ~2  -  2  ay  =  0,  (2) 
the  elements  of  the  cylin- 
der being  parallel  to  OX. 
To  find  the  projection 
of  the  required  area  on 
the  plane  XOY  it  is 
necessary     to     find     the 

projection  on  that   plane   of   the    line   of    intersection   of   (1)   and   (2). 
Hence  (§  lo'.t)  we  must  eliminate    :   from  (1)  and   (•_').     The   result    is 


1  a-  =  0. 


(3) 


From  (2), 


0,         - 


^(iHtr=^§ 


Jfl!  = 


and 


dA 


adydx 

\  •_'  a~y  -  f 

itive   side  of   t!u 


Since  the  area  on  the  positive  side  of  the  plane  XOY  is  symmetrica] 

■with  respect  to  the  plane    YOZ,  it   is  twice  the  area  on  the   positive  side  of 
the  latter  plane.    Hence  we  may  find  this  latter  area  and  multiply  by  2. 

If  the  first  integration  is  with  reaped  to  x,  the  lower  limit  is  evidently 
0  and  the  upper  limit,  found  from  (:'>),  is  V4 a2  —  2 ay.  For  the  final 
integration  with  respect  to  y  the  limits  are  <)  and  2a,  the  latter  being 
found  from  (3). 

adydx 


Therefore    .1 


Jn      Jo 


Vi* 


8  <fi. 


As  an  equal  area  is  intercepted  on  the  negative  .side  of  the  plane  XOY,  the 
above  result  must  be  multiplied  by  2.   Hence  the  total  required  area  is  16  a-. 

The  evaluation  of  (4)  may  sometimes  be  simplified  by  trans- 
forming to  polar  coordinates  in  the  plane  XOY. 


384 


MULTIPLE  INTEGRALS 


Ex.  3.  Find  the  area  of  the  sphere  x"  +  f  +  z"  =  a2  included  in  a 
cylinder  having  its  elements  parallel  to  OZ  and  one  loop  of  the  curve 
r  =  a  cos  2d  (fig.  225)  in  the  plane  XOY  as  its  directrix. 

Proceeding  as  in  Ex.  1,  we  find  the  integrand  ■ —  Trans- 

V«2  —  x2  —  if 

forming  this  integrand  to  polar  coordinates,  and  multiplying  by  rdrdO, 

we  have 

,  .         ardrdd 
(I  A  — 


Fig.  225 


V«2  -  r2 

It  is  evident  that  the  required  area  q 
is  twice  the  area  cut  out  of  the  sphere 
on  one  side  of  the  plane  XOY,  and 
that  this  latter  area  is  twice  the  area 
over  the  half  of  the  loop  of  the  curve 
r  —  a  cos  2  6  which  is  in  the  first 
quadrant. 

Hence  we  integrate  first  with  respect  to  r  from  0  to  a  cos  2  6,  and  then 

integrate  with  respect  to  6  from  0  to  —  >  the  limits  of  integration  all  being 

determined  from  the  equation  r  —  a  cos  2  6.    This  integral  we  multiply  by 
4  to  obtain  the  required  area. 

Jo    Jo  Va2  -  >•-' 

=  4rr  r¥(l-sin2  0W0 

Jo 


Therefore 


If  the  required  area  is  projected  on  the  plane  YOZ,  we  have 

(5) 


'-J5M 


-'l)  +  (i" 


where  the  integration  extends  over  the  projection  of  the  area 
on  the  plane  YOZ;  and  if  the  required  area  is  projected  on  the 
plane  XOZ,  we  have 


fU 


^B+ay** 


(6) 


where  the  integration  extends  over  the  projection  of  the  area 
on  the  plane  XOZ. 


TRIPLE  INTEGRALS 


385 


181.  Triple  integrals.  1.  Rectangular  coordinates.  Let  any 
volume  (fig.  226)  be  divided  into  rectangular  parallelepipeds 
of  volume  AxAyAz  by  planes  parallel  respectively  to  the  coor- 
dinate planes,  some  of  the  parallelepipeds  extending  outside 
the  volume  in  a  manner  similar  to  that  in  which  the  rectangles 
in  §  174  extend  outside  the  area.  Let  (a;,.,  yp  z^)  be  a  point 
of  intersection  of  any  three  of  these  planes  and  form  the  sum 


as  in  §  174.  Then  the  limit  of  this  sum  as  n,  m,  and  p  increase 
indefinitely,  while  Ax  =  0,  Ay  =  0,  Az  =  0,  so  as  to  include  all 
points  of  the  volume,  is  called  the  triple  integral  of  ,/'(./-,  y,  z) 
throughout  the  volume.  It  is 
denoted  by  the  symbol 


Iff 


/'(./■.  y,  z)dxdydz,    (1) 


the  limits  remaining  to  be  substi 
tuted.  If  the  summation  is  made 
first  with  respect  to  z,  x  and  y 
remaining  constant,  the  result  is 
to  extend  the  integration  through- 
out a  column  of  cross  section 
Ax  Ay  ;  if  next  ./•  remains  constant 
and  y  varies,  the   integration   is 

extended  so  as  to  combine  the  columns  into  slices;  and,  finally, 
as  x  varies,  the  slices  are  combined  so  as  to  complete  the 
integration  throughout  the  volume. 

The  volume  of  the  parallelepiped  with  edges  d.r,  dy,  dz  is  the 
element  of  volume  c/F,  and  hence 


Fig.  tk\ 


dV=  dxdydz. 


(2) 


2.  Cylindrical  coordinates.  If  the  x  and  the  y  of  the  rectan- 
gular coordinates  are  replaced  by  polar  coordinates  r  and  6  in 
the  plane  XOY,  and  the  z  coordinate  is  retained  with  its  original 


386 


MULTIPLE  INTEGRALS 


Fig. 


significance,  the  new  coordinates  r,  0,  and  z  are  called  cylindri- 
cal coordinates.  The  formulas  connecting  the  two  systems  of 
coordinates  are  evidently 

x  =  r  cos  0,     y  —  r  sin  0,     z  —  z. 

Turning  to  fig.  227,  we  see  that  z  =  z1 
determines  a  plane  parallel  to  the  plane 
XOY,  that  0  =  Qx  determines  a  plane 
MONP,  passing  through  OZ  and  making 
an  angle  01  with  the  plane  XOZ,  and 
that  r  —  rx  determines  a  right  circular 
cylinder  with  radius  r1  and  OZ  as  its 
axis.  These  three  surfaces  intersect  at 
the  point  P. 

The  element  of  volume  in  cylindrical  coordinates  (fig.  228) 
is  the  volume  bounded  by  two  cylin- 
ders of  radii  r  and  r  +  Ar,  two  planes 
corresponding  to  z  and  z  +  Az,  and 
two  planes  corresponding  to  0  and 
0  +  A0.  It  is  accordingly  a  cylinder 
with  its  altitude  equal  to  Az  and  the 
area  of  its  base  approximately  equal 
to  rA0Ar  (§  176).  Hence,  in  cylin- 
drical coordinates, 

dV=rdrdddz,  (3)     /  Fig.  228 

and  the  triple  definite  integral  in  cylindrical  coordinates  is 


Iff 


f(r,  0,  z)rdrd0dz,         (4) 


the  limits  remaining  to  be  substituted. 
3.  Polar  coordinates.  In  fig.  229  the 
cylindrical  coordinates  of  P  are  OM=  r, 
MP  =  z,  and  ZLOM=  0.  If  instead  of 
placing  OM  =  r  we  place  OP  =  r,  and 
denote  the  angle  NOP  by  (/>,  we  shall 
have  r,  </>,  and  0  as  the  polar  coordi- 
nates of  P.    Then,  since  0N=  OP  cos  <f> 


TRIPLE  INTEGRALS 


387 


and  0M=  OB  sin  <f>,  the  following  equations  evidently  express 
the  connection  between  the  rectangular  and  the  polar  coordi- 
nates of  P: 

z  =  r  cos  (f>,  x  =  r  sin  <f>  cos  6,  y  =  r  sin  (f>  sin  6. 

The  polar  coordinates  of  a  point  determine  three  surfaces 
which  intersect  at  the  point.  For  6  =  6^  determines  a  plane 
(fig.  230)  through  OZ,  making 
the  angle  0  with  the  plane  XOZ\ 
4>  =  (f>l  determines  a  cone  of  revo- 
lution, the  axis  and  the  vertical 
angle  of  which  are  respectively 
OZ  and  2^;  and  r  =  r^  drier- 
mines  a  sphere  with  its  center 
at  0  and  radius  r . 

The  element  of  volume  in 
polar  coordinates  (fig.  231)  is 
the  volume  hounded  by  two 
spheres  of  radii  r  and  r  +  Ar, 
two  conical  surfaces  corresponding  to  (f>  and  (f>  +  A$,  and  two 
planes  corresponding  to  6  and  6  +  A$.  The  volume  of  the 
spherical  pyramid  O-ABCD  is 
equal  to  the  area  of  its  base 
ABCD  multiplied  by  one  third 
of  its  altitude  r.  To  find  the 
area  of  ABCD  we  note  first  that 
the  area  of  the  zone  formed  by 
completing  the  arcs  AD  and  BC 
is  equal  to  its  altitude,  rcos$  — 
r  cos  ((f)  +  A<£),  multiplied  by  2irr. 
Also  the  area  of  ABCD  is  to  the 
area  of  the  zone  as  the  angle  Ad 

is   tO    2  7T. 


Hence  area  ABCD 

and  vol  O-ABCD  ■■ 

Similarly,  vol  0~EFGII=  l(r+  A/-) 3A0 [cos <f>- cos (<f>  +  A<£)]. 


rA6  [r  cos  (f)  —  r  cos  ((f)  +  A$)] 
I  r3Ad  [cos  <f>  -  cos  ((f)  +  A(£)]. 


388  MULTIPLE  INTEGRALS 

Therefore  vol  ABCDEFGH  =  J  [(/•  +  Ar)3  -  r3]  A0  [cos  <f> 

-eos(<£+ A<£)]. 
i  [(r  +  A;-)3  -  >•']  =  '/2Ar  +  rAr*  +  I  AT-3 
=  (r2  +  e1)Ar. 
cos  (/>  —  cos  ((£  +  A<£)  =  —  A  cos  c£ 

=  (sin  <£  +  e2)  A<£.  (§77) 

Hence        vol  ABCDEFGH  =  (f  sin  <f>  +  e8)  A>- A0  A<£, 

which    differs  from    r2  sin  0  ArA0A</>  by  an    infinitesimal  of   a 
higher  order. 

Accordingly  we  let  dV  =  r2  sin  <f>drd<j)d6,  (5) 

and  the  triple  integral  in  polar  coordinates  is 


W> 


/(r,  (/>,  0)  r2  sin  <f>  dr  d<$>  dd,  (6) 


the  limits  remaining  to  be  substituted. 

It  is  to  be  noted  that  dV  is  equal  to  the  product  of  the 
three  dimensions  AB,  AD,  and  A£J,  which  are  respectively  rd<f>, 
r  sin  0  dO,  and  dr. 

182.  Change  of  coordinates.    When  a  double  integral  is  given 

in  the  form  jjf(x,  y)dxdy,  where  the  limits  are  to  be  substi- 
tuted so  as  to  cover  a  given  area,  it  may  be  easier  to  determine  the 
value  of  the  integral  if  the  rectangular  coordinates  are  replaced 
by  polar  coordinates.  Then  f(x,  y)  becomes  f(r  cos  6,  r  sin  0), 
i.e.  a  function  of  r  and  6.  As  the  other  factor,  dxdy,  indicates 
the  element  of  area,  we  may  replace  dxdy  by  rdrdd.  These  two 
elements  of  area  are  not  equivalent,  but  the  two  integrals  are 
nevertheless  equivalent,  provided  the  limits  of  integration  in  each 
system  of  coordinates  are  taken  so  as  to  cover  the  same  area. 
In  like  manner  the  three  triple  integrals 


HP 
Iff 

Iff 


f(x,  y,  z)dxdydz, 

f(r  cos  6,  r  sin  6,  z)rdrdddz, 

f(r  sin  <f>  cos  0,  r  sin  $  sin  0,  r  cos  (/>)  r2  sin  <f>drd<f>d6 


VOLUME 


389 


are  equivalent  when  the  integration  is  taken  over  the  same  volume 
in  all  three  and  the  limits  are  so  taken  in  each  as  to  include  the 
total  volume  to  be  considered. 

183.  Volume.  In  §  181  we  found  expressions  for  the  element 
of  volume  in  rectangular,  in  cylindrical,  and  in  polar  coordinates. 
The  volume  of  a  solid  bounded  by  any  .surfaces  will  be  the  limit 
of  the  sum  of  these  elements  as  their  number  increases  indefi- 
nitely while  their  magnitudes  approach  the  limit  zero.  It  will 
accordingly  be  expressed  as  a  triple  integral. 


Find  the  volume  bounded  by  the  ellipsoid 


/,- 


=  1. 


From  symmetry  (fig.  232)  it  is  evident  that  the  required  volume  is 
eight  times  the  volume  in  the  first  octant  b6unded  by  the  Burface  and  the 
coordinate  planes. 

In  summing  up  the  rectangu- 
lar parallelepipeds  dxdydz  to 
form  a  prism  with  edges  paral- 
lel to  OZ,  the  limits  for  z  are  0 


and 


V.      x2      y* 


the    latter 


being  found  from  the  equation 

of  the  ellipsoid. 

Summing  up  next  with  respect 
to  y,  to  obtain  the  volume  of  a 
slice,  we  have  0  as  the  lower  limit 


of  y,  and  6 a/I 


V 


Fi 

-  as  the  upper  limit.    This  latte 
1,  found  by  letting 


■  VI 


limit  is  determined 
:  =  0  in  the  equation 


by  solving  the  equation  —  4-  '— 

of  the  ellipsoid;  for  it  is  in  the  plane  z  =  0  that  the  ellipsoid  has  the 
greatest  extension  in  the  direction  <>)',  corresponding  to  any  value  of  x. 
Finally,  the  limits  for  x  are  evidently  0  and  o. 


Therefore 


Jo    t/0  Jo 


^dxdydz 


dxdy 


390  MULTIPLE  INTEGEALS 

It  is  to  be  noted  that  the  first  integration,  when  rectangular 
coordinates  are  used,  leads  to  an  integral  of  the  form 


//< 


(Zo-z^dxdy, 

where  z2  and  zy  are  found  from  the  equations  of  the  boundary 
surfaces.  It  follows  that  many  volumes  may  be  found  as  easily 
by  double  as  by  triple  integration. 

In  particular,  if  zx=  0,  the  volume  is  the  one  graphically  repre- 
senting the  double  integral  (§  174). 

Ex.  2.  Find  the  volume  bounded  by  the  surface  z  =  ae-^  +  v2)  and  the 
plane  z  =  0. 

To  determine  this  volume  it  will  be  advantageous  to  use  cylindrical 
coordinates.  Then  the  equation  of  the  surface  becomes  z  =  ae~r',  and  the 
element  of  volume  is  (§  181)  rdrdOdz. 

Integrating  first  with  respect  to  z,  we  have  as  the  limits  of  integration 
0  and  ae~  r2.  If  we  integrate  next  with  respect  to  r,  the  limits  are  0  and  go, 
for  in  the  plane  z  =  0,  r  =  cc,  and  as  ~  increases  the  value  of  r  decreases 
toward  zero  as  a  limit.  For  the  final  integration  with  respect  to  6  the 
limits  are  0  and  2tt. 


Therefore 


V=   f"K  f°°fae   '  rdOdrdz 

Jo      Jo    Jo 

=  a  f2n  f"re-r\l9dr 
Jo     Jo 


In  the  same  way  that  the  computation  of  the  volume  in  Ex.  2 
has  been  simplified  by  the  use  of  cylindrical  coordinates,  the 
computation  of  a  volume  may  be  simplified  by  a  change  to  polar 
coordinates ;  and  the  student  should  always  keep  in  mind  the 
possible  advantage  of  such  a  change. 

184.  Moment  of  inertia  of  a  solid.  Following  the  method 
of  §  178  we  divide  the  volume  of  the  solid  into  n  elements  Ay 
and  multiply  each  element  by  its  density  p.  Then,  if  Rt  is  the 
distance  of  any  point  of  the  ith.  element  from  the  axis  about 
which  the  moment  of  inertia  is  to  be  taken,  we  may  take  It'fpAv 


MOMENT  OF  INERTIA  391 

as  the  moment  of   inertia  of  that  element.    If  /  denotes  the 

moment  of  inertia  of  the  solid,   Vi?.2pAv  is  an  approximate 

expression  for  /.  Finally,  if  we  let  n  =  oo  and  at  the  same  time 
let  each  element  of  volume  approach  zero  as  a  limit,  we  have 

1=  Limit  2}  Hfpkv  =  J  Kydv, 

where  B,  p,  and  civ  are  to  be  expressed  in  terms  of  the  same 
variables  and  the  proper  limits  of  integration  substituted.  In 
particular,  if  civ  is  replaced  by  any  one  of  the  three  elements 
of  volume  determined  in  §  181,  the  integral  becomes  a  triple 
integral. 

Ex.  Find  the  moment  of  inertia  of  a  sphere  of  radius  a  about  a  diameter 
if  the  density  varies  directly  as  the  square  of  the  distance  from  the  diameter 
about  which  the  moment  of  inertia  is  to  be  taken. 

We  shall  take  the  center  of  the  sphere  as  the  origin  of  coordinates,  and 
the  diameter  about  which  the  moment  is  to  be  taken  as  the  axis  of  r.  The 
problem  will  then  be  most  easily  solved  by  using  cylindrical  coordinates. 

The  equation  of  the  sphere  will  be  r2  +  «2  =  a2,  and  dv  =  rdrd6dz) 
R  =  r,  and  p  =  kr'2,  so  that  we  have  to  find  the  value  of  the  triple  integral 

*///•*»**■ 

Since   the   solid   is   symmetrical   with    respecl    to   the   plane   2  =  0,  we 

shall   take  0  and  \f<r  —  /■-  as  the   limits  of   integration   with    respect  to  z, 

the  latter  limit  being  found  from  the  equation  of  the  sphere,  and  double 
the  result. 

Integrating  next  with  respect  to  r,  we  have  the  limits  0  and  r/,  thereby 
finding  the  moment  of  a  sector  of  the  sphere.  To  include  all  the  sectors, 
we  have  to  take  0  and  2  ir  as  the  limits  of  0  in  the  last  integration. 

Therefore  I=2lC'U  C"  C        '  V<  ,16  dr  dz. 

Jo        Jo    Jo 

As  a  result  of  the  first  integration, 

/  =  2  /.'  f ' "  f  "  r'  N  '"'-  -~7-  dd  dr. 
Jo     Jo 

After  the  second  integration, 

I =lo%^f^d6, 
and,  finally,  /=  ffig  Ittu1. 


392  MULTIPLE  INTEGRALS 

185.  Center  of  gravity  of  a  solid.  The  center  of  gravity  of  a 
solid  has  three  coordinates,  x,  y,  z,  which  are  defined  by  the 
equations 

I  xdm  j  ydm  j  zdm 


I  dm  j  dm  j  i 


hit 


where  dm  is  an  element  of  mass  of  the  solid  and  r,  y,  and  z  are 
the  coordinates  of  the  point  at  which  the  element  dm  may  be 
regarded  as  concentrated.  The  derivation  of  these  formulas  is 
the  same  as  that  in  §  137. 

When  dm  is  expressed  in  terms  of  space  coordinates,  the 
integrals  become  triple  integrals,  and  the  limits  of  integration 
are  to  be  substituted  so  as  to  include  the  whole  solid. 

The  denominator  of  each  of  the  preceding  fractions  is  evi- 
dently M,  the  mass  of  the  body. 

Ex.  Find  the  center  of  gravity  of  a  body  of  uniform  density,  bounded 
by  one  nappe  of  a  right  circular  cone  of  vertical  angle  2  a  and  a  sphere  of 
radius  a,  the  center  of  the  sphere  being  at  the  vertex  of  the  cone. 

If  the  center  of  the  sphere  is  taken  as  the  origin  of  coordinates  and  the 
axis  of  the  cone  as  the  axis  of  z,  it  is  evident  from  the  symmetry  of  the 
solid  that  x  —  y  =  0.  To  find  z,  we  shall  use  polar  coordinates,  the  equations 
of  the  sphere  and  the  cone  being  respectively  r  =  a  and  <f>  =  a. 

f  ^  f  f°r  cos  <£  •  r2  sin  (fydOd^dr 

Then  ~  =  Jo     JoJ\ 

f'n  f"  f  V  sin  tf> (Id d<l>dr 
Jo     Jo  Jo 

The  denominator  is  the  volume  of  a  spherical  cone  the  base  of  which 
is  a  zone  of  one  base  with  altitude  a  (1  —  cos  ex)  ;  therefore  its  volume 
equals  §  ira3  (1  —  cos  a). 

f       f     f    >'3  cos  $  sin  <jid9d(f>dr  =  \  a*  C       C    cos  <£  s'mcf}d6dcf> 

=  1  a4(l-cos2«)  f2nd9 
Jo 

=  1 :7ra4(l  —  cos2  a:). 
Therefore  z  =  j}  (1  +  cos  a)  a. 


ATTRACTION 


186.  Attraction.    The  formula 

'cos  6  dm 


P 


ir 


(§  138) 


for  the  component  of  attraction  in  the  direction  OX  is  entirely 
general.  Similar  formulas  for  the  components  in  the  direc- 
tions OY  and  OZ  may  be  deduced.  The  application  of  these 
formulas  requires  us  to  express 
B,  cos  6,  and  dm  in  terms  of 
the  same  variables,  and  to  substi- 
tute limits  of  integration  so  as 
to  include  the  whole  of  the 
attracting  mass.  In  general,  the 
integral,  after  the  substitution  of 
the  variables,  will  be  a  double 
or  triple  integral. 

Ex.  Find  the  attraction  due  to  a 
homogeneous  circular  cylinder  of 
density  p,  of  height  h,  and  radius 
of  cross  section  a,  on  a  particle  in 
the  line  of  the  axis  of  the  cylinder 
at  a  distance  l>  units  from  one  end 
of  the  cylinder. 

Take  the  particle  at  the  origin  of 
coordinates  (fig.  233),  and  the  axis  of 
the  cylinder  as  ()Z.  Using  cylindrical 
coordinates,  we  have  dm  =  prdrdddz 
and   R  =  Vz~  +  r2. 

From  tlie  symmetry  of  the   figure  the 


Fig.  233 


■sultant  components  of  attrac- 


tion in  the  directions  OX  and  O  Y  are  zero,  and  cos  6  = 


resultant  component  in  the  direction  OZ. 
1    represent 

X2  7T     r%  a     r\b 
i  JO     Jb 


Vr2 


for  the 


Therefore,   letting   A    represent  the   component  in   the   direction   OZ, 
+  h 


we  have 


dddrdz, 


(c2  +  r2)2 
where  the  limits  of  integration  are  evident  from  fig.  233. 

A  =  p  f*'  ["I        r         -  r  )  dOdr 

Jo     Jo   VVi2  +  r2       V(h  +  h)2  +  /•-/ 

=  P  f2w(k  +  V62  +  a*  -  V(/>  +  ?>y2  +  «2)  <io 


S¥ 


V(/;  +  ],y  +  a2). 


394  MULTIPLE  INTEGRALS 

PROBLEMS 
Find  the  values  of  the  following  integrals : 

iff   %dxdy.  4.   J      I    log^dxdy 

2.    C^f^xcosydxdy.  5"  f  £         {*  +  V)*V** 


Jo     Jo 


6.  Express  in  two  ways  the  inte- 
gral off(x,  y)  over  the  smaller  area 

sin  (x  +  //)  dx  dy.    bounded  by  the  curves  x  +  V  =  2  a 
and  (x  —  a)'1  +  y2  —  a2. 


Find  the  values  of  the  following  integrals : 
7.    /      /  rdtfdr.  8.    /       /         t*mn*$dOdr. 

Jo     Jo  J-  1 17 o 

9.  Find  the  integral  of  r  over  one  loop  of  the  curve  r  =  a  sin  2  0. 

10.  Find  the  integral  of  r  over  the  area  bounded  by  the  initial 
line  and  the  curves  r  =  a  and  r  —  a(l  -4-  cos  0). 

11.  Find  the  area  bounded  by  the  curves  y  =  x%  and  ?/  =  2  —  x2. 

12.  Find  the  area  bounded  by  the  hyperbola  xy  =  4  and  the  line 
x  +  y  -  5  =  0. 

13.  Find  the  area  bounded  by  the  confocal  parabolas  if  =  4  ax 
+  4  aa,  v/  =  -  4  Ax  +  4  h\ 

14.  Find  the  area  of  the  loop  of  the  curve  (a;  +  yf  =  if  (//  +1). 

15.  Find  the  area  bounded  by  the  curves  x2  +  if  =  25,  3  y2  =  16  x, 
3x2  =  16y. 

16.  Find  each  of  the  areas  bounded  by  the  circle  x2  -f-  if  —  5  a2 

and  the  witch  ?/  =  — — — — -• 
x-  +  4  ar 

17.  Find  the  area  bounded  by  the  circles  r  =  a  cos  6,  r  =  a  sin  6. 

18.  Find  the  area  cut  off  from  a  loop  of  the  curve  r  =  a  cos  2  0 
by  the  curve  r  =  —  • 

19.  Find  the  area  cut  off  from  the  lemniscate  r2  =  2«2cos  2  0  by 
the  straight  line  r  cos  0  =  -jr-  a. 


PROBLEMS  395 

20.  Find  the  area  bounded  by  the  limacon  r  =  2  cos  6  +  3  and 
the  circle  r  =  2  cos  6. 

21.  Find  the  area  which  is  outside  the  circle  r=a  and  inside 
the  cardioid  r  =  a(l  +  cos  6). 

22.  Find  the  area  in  the  first  quadrant  bounded  by  the  circle 
r  —  2a  sin  $  and  the  lemniscate  r2  =  2  a2  cos  2  0. 

23.  Find  the  moment  of  inertia  of  the  area  bounded  by  the  hyper- 
bola xy  =  4  and  the  line  x  +  y  —  5  =  0  about  an  axis  perpendicular 
to  its  plane  at  O. 

24.  Find  the  moment  of  inertia  of  the  area  bounded  by  the  curves 
y  =  /2,  y=2  —  x'2  about  an  axis  perpendicular  to  its  plane  at  0. 

25.  Find  the  moment  of  inertia  about  OY of  the  area  bounded  by 
OK  and  the  parabola  y2  =  1  —  x. 

26.  Find  the  moment  of  inertia  aboul  an  axis  through  0  perpen- 
dicular to  the  coordinate  plane  of  thai  pari  of  the  first  quadrant 
included  between  the  first  two  successive  coils  of  the  spiral  /•  =  e°*. 

27.  Find  the  moment  of  inertia  of  the  entire  area  bounded  by 
the  curve  r  —  a3 sin 30  about  an  axis  perpendicular  to  its  plane  at 

the  pole. 

28.  Find  the  moment  of  inertia  of  the  area  of  one  loop  of  the  lem- 
niscate j-2  =  2  «2  cos  2  9  about  an  axis  perpendicular  to  its  plane  at 
the  pole. 

29.  Find  the  moment  of  inertia  of  the  total  area  bounded  by  tin- 
curve  1*  =  a2  sin  6  about  an  axis  in  its  plane  perpendicular  to  the 
initial  line  at  the  pole. 

30.  Find  the  moment  of  inertia  about  OX  of  the  area  bounded  by 
the  parabolas  if  —  4  ax  +  4  a3,  y2  =  —  4  />./■  +  4  U2. 

31.  Find  the  moment  of  inertia  about  OF  of  the  area  of  the  loop 
of  the  curve  y2  =  x2(2  —  x). 

32.  Find  the  moment  of  inertia  of  the  area  of  the  cardioid 
r  —  a  (1  +  cos  6)  about  an  axis  perpendicular  to  its  plane  at  the  pole. 

33.  Find  the  moment  of  inertia  of  the  area  of  a  circle  of  radius  a 
about  an  axis  perpendicular  to  the  plane  of  the  circle  at  any  point 
on  its  circumference. 

34.  Find  the  moment  of  inertia  about  its  base  of  the  area  of  a 
parabolic  segment  of  height  h  and  base  2  a. 


396  MULTIPLE  INTEGRALS 

35.  Find  the  moment  of  inertia  about  OX  of  the  area  bounded  on 
the  left  by  an  arc  of  the  curve  y2  =  ax  -4-  as  and  on  the  right  by  an 
arc  of  the  curve  x"  +  if  =  a'1. 

36.  Find  the  moment  of  inertia  of  the  area  of  one  loop  of  the 
lemniscate  r*  =  2  a2  cos  2  6  about  an  axis  in  its  plane  perpendicular 
to  the  initial  line  at  the  pole. 

37.  Find  the  moment  of  inertia  about  the  initial  line  as  an  axis 
of  the  area  of  the  cardioid  r  =  a  (cos  6  -\- 1)  above  the  initial  line. 

38.  Find  the  moment  of  inertia  of  the  area  bounded  by  a  semi- 
circle of  radius  a  and  the  corresponding  diameter,  about  the  tangent 
parallel  to  the  diameter. 

39.  Find  the  moment  of  inertia  of  the  area  of  a  loop  of  the  curve 
v  =  a  cos  2  6  about  the  axis  of  the  loop  as  an  axis. 

40.  Find  the  moment  of  inertia  of  the  area  of  the  circle  r  =  a 
which  is  not  included  in  the  curve  r  =  a  sin  2  6  about  an  axis 
perpendicular  to  its  plane  at  the  pole. 

41.  Determine  the  center  of  gravity  of  the  half  of  a  parabolic 
segment  of  altitude  9  in.  and  of  base  12  in.  formed  by  drawing  a 
straight  line  from  the  vertex  of  the  segment  to  the  middle  point 
of  its  base. 

42.  Find  the  center  of  gravity  of  a  lamina  in  the  form  of  a 
parabolic  segment  of  altitude  7  in.  and  of  base  28  in.  if  the  density 
at  any  point  of  the  lamina  is  directly  proportional  to  its  distance 
from  the  axis  of  the  lamina. 

43.  Find  the  center  of  gravity  of  the  area  of  a  loop  of  the  curve 
ay  =  «V  -  x6. 

44.  Find  the  center  of  gravity  of  the  area  bounded  by  the  parabola 
x*  -4-  ?/2  =  a*  and  the  circle  x2  +  if  =  a2. 

45.  Find  the  center  of  gravity  of  the  area  bounded  by  the  cardioid 
r  =  a  (cos  6  +  1). 

46.  Find  the  center  of  gravity  of  the  area  bounded  by  the  parabola 

x2  =  4  aii  and  the  witch  y  =  -= : — -0  • 

J  J       x1  +  4  a2 

47.  A  plate  is  in  the  form  of  a  sector  of  a  circle  of  radius  a,  the 
angle  of  the  sector  being  2  a.  If  the  thickness  varies  directly  as  the 
distance  from  the  center,  find  its  center  of  gravity. 


PROBLEMS  397 

48.  Find  .the  center  of  gravity  of  the  area  in  the  first  quadrant 
bounded  by  the  curve  x$  +  //'•  =  "•'■  and  the  line  x  4-  y  =  a. 

49.  The  density  at  any  point  of  a  lamina  in  the  form  of  a  loop 
of  the  curve  r  ==  a  cos  2  8  is  directly  proportional  to  its  distance 
from  the  point  of  the  loop.    Determine  its  center  of  gravity. 

50.  Find  the  center  of  gravity  of  the  area  bounded  by  the  limacon 
r  =  2  cos  $  4-  3. 

51.  Find  the  center  of  gravity  of  the  area  bounded  by  the  curve 
v  =  a  sin  -  as  $  changes  from  0  to  2  it. 

52.  Find  the  center  of  gravity  of  the  area  bounded  by  the  cissoid 

x3 
f  = and  its  asymptote. 

53.  Find  the  center  of  gravity  of  the  area  cut  off  from  the  lemnis- 

V3 

cate  r2  =  2  a2  cos  2  6  by  the  straight  line  rcos0=  -jr-  a. 

■'■'-       'r 

54.  From  a  homogeneous  elliptic  plate,  —,  +  ',.,  —  F  is  cut  a  circular 

plate  of  radius  /•(/•  <  -J  with  center  at  f-j  OJ.    Find  the  center  of 
gravity  of  the  part  left. 

55.  Find  the  area  of  the  surface  cut  from  the]  >aral  ><  >loid  //"+  v-  =  4  <is 
by  the  cylinder  //-  =  ax  and  the  plane  ./•  =  .'!  a. 

56.  Find  the  area,  of  the  surface  of  the  cone  ./-  4-  //-  —  4  ::2  =  0 
cut  out  by  the  cylinder  .r  +  //-  —  4./-  =  0. 

57.  Find  the  area  of  the  surface  cut  from  the  cylinder  ./•-  4-  //  =  a2 
by  the  cylinder  if  +  --  =  a'1. 

58.  Find  the  area  of  the  surface  of  a  sphere  of  radius  a  inter- 
cepted by  a  right  circular  cylinder  of  radius  \  a,  if  an  element  of 
the  cylinder  passes  through  the  center  of  the  sphere. 

59.  Find  the  area  of  the  sphere  x2  4-  //  +  z2  =  a2  included  in  the 
cylinder  with  elements  parallel  to  OZ  and  having  for  its  directrix  in 
the  plane  XOY  a  single  loop  of  the  curve  r  =  a  cos  30. 

60.  Find  the  area  of  the  surface  of  the  cylinder  x2  4-  if  —  2  ax  =  0 
bounded  by  the  plane  ZOFand  a  right  circular  cone  having  its  vertex 
at  0,  its  axis  along  OZ,  and  its  vertical  angle  equal  to  90°. 


398  MULTIPLE  INTEGRALS 

61.  Find  the  area  of  the  paraboloid  x2  4-  y2  =  2  az  included  in  the 
cylinder  with  elements  parallel  to  OZ  and  having  for  its  directrix  in 
the  plane  XOY  one  loop  of  the  curve  r2  =  a2  sin  2  0. 

62.  Find  the  area  of  the  surface  z  —  xy  included  in  the  cylinder 

X1  yz 

63.  Find  the  area  of  that  part  of  the  surface  z  =  —  the  projec- 
tion  of  which  on  the  plane  XOY  is  bounded  by  the  curve  r2  =  a2  cos  0. 

64.  Find  the  area  of  the  surface  of  the  cylinder  x2  +  y2  —  2  ax  =  0 
included  in  the  cone  x2  —  y2  +  2  z2  =  0. 

65.  Find  the  area  of  the  sphere  x2  +  y2  4-  z2  =  4«2  bounded  by 
the  intersection  of  the  sphere  and  the  right  cylinder  the  elements 
of  which  are  parallel  to  OZ  and  the  directrix  of  which  is  the  cardioid 
r  =  a  (cos  0  +  1)  in  the  plane  XOY. 

66.  Find  the  area  of  the  surface  of  the  sphere  (x  —  a)2+y2  +  z2  =  a2 
included  in  one  nappe  of  the  cone  x2  4-  y2  —  z2  =  0. 

Find  the  values  of  the  following  integrals  : 
dxdydz 


n*8   r^^-y2    dxdy< 
1  **~f 


67. 

V^T^i    /^V„2_x2_j/2         yz dxdydz 


68 


11      1        ■* 


J  ex  +  y  +  zdxdydz. 

Jo 

70.     /       I       /  r3drd$dz. 

Jo     Jo    Jo 

,Vf7T72 
7-^  rzdddrdz 


J  -iJa&mBJO 


72. 


{a2  -  r2  +  z2y 
•dz 


n  fcme  fdOdra 
Jo     J,         1        »V 

73.     I       I       I        r  sin2  <f>  cos  <f>  cos  0  d6d(f>  dr. 

%J0      i/0      t/asin0 

Xtt      /-» 2  tt      /»a  cos  0 
/         I  r  sin3  <f>d<f>d0 dr. 

Jo      Jo 


PEOBLEMS  399 

75.  Find  the  volume  bounded  by  the  surface  x^  +  y^  +  z^  =  a? 
and  the  coordinate  planes. 

76.  Find  the  volume  of  a  cylindrical  column  bounded  by  the 
surfaces  y  =  x2,  x  =  if,   z  =  0,  z  =  12  +  y  —  x2. 

77.  Find  the  volume  bounded  by  the  plane  z  =  0  and  the 
cylinders  x2  +  y2  =  a2,  f  =  a2  —  az. 

78.  Find  the  volume  bounded  by  the  surfaces  r2  =  bz,  z  =  0, 
r  —  a  cos  0. 

79.  Find  the  volume  bounded  by  the  sphere  x2  -\-  f  +  z2  =  5  and 
the  paraboloid  x2  +  if  —  4  z. 

80.  Find  the  volume  bounded  by  the  cylinder  z2  =  x  -\-  y  and  the 
planes  x  =  0,  y  =  0,  z  =  4. 

81.  Find  the  volume  of  the  paraboloid  y2  +  z2  =  2  x  cut  off  by 
the  plane  y  =  x  —  1. 

82.  Find  the  volume  bounded  by  a  sphere  of  radios  "  and  a  right 
circular  cone,  the  axis  of  the  cone  coinciding  with  a  diameter  of  the 
sphere,  the  vertex  being  at  an  end  of  the  diameter  and  the  vertical 
angle  of  the  cone  being  90°. 

83.  Find  the  total  volume  bounded  by  the  surface  (x2  ■+■  y2  +  z2)3 
=  27  asxyz.    (Change  to  polar  coordinates.) 

84.  Find  the  volume  bounded  by  the  plane  XOY,  the  cylinder 
.'""  +  if"  —  2aa!  =  0,  and  the  right  circular  cone  having  its  vertex  at 
<>,  its  axis  coincident  with  UZ,  and  its  vertical  angle  equal  to  90°. 

x2       i/2      z* 

85.  Find  the  total  volume  bounded  by  the  surface  —2  +  'j^  +  —  =  1. 

86.  Find  the  volume  bounded  below  by  the  paraboloid  x2  +  y2  =  az 
and  above  by  the  sphere  x2  +  if  +  z2  —  2  az  =  0. 

87 .  Find  the  volume  bounded  by  the  surface  b2z2  =  if  (a2  —  x2) 
and  the  planes  y  —  0,  y  =  b. 

88.  Find  the  volume  cut  from  a  sphere  of  radius  a,  by  a  right 
circular  cylinder  of  radius  ->  one  element  of  the  cylinder  passing 
through  the  center  of  the  sphere. 

89.  Find  the  total  volume  bounded  by  the  surface  (x2  +  y2  +  z2)2 
=  axyz. 


400  MULTIPLE  INTEGRALS 

90.  Kind  the  volume  in  the  first  octant  bounded  by  the  surfaces 
*  =  (a  +  !/f,  x1  +  if  =  «a- 

91.  Find  the  volume  of  the  sphere  j'2  +  //2  +  z2  =  a?  included  in 
a  cylinder  with  elements  parallel  to  OZ,  and  having  for  its  directrix 
in  the  plane  XOY  one  loop  of  the  curve  r2  =  a2  cos  2  0. 

92.  Find  the  volume  bounded  by  the  surfaces  az  =  xy,  x  +  y  +  z 
=  a,  z  =  0. 

93.  Find  the  total  volume  which  is  bounded  by  the  surface  x*  +  //■'> 

+  ~i  =  J. 

94.  Find  the  total  volume  which  is  bounded  by  the  surface  r2  -f-  z" 
=  2  ar  cos  2  6. 

95.  Find  the  moment  of  inertia  about  its  axis  of  a  hollow  right 
circular  cylinder  of  mass  M,  the  inner  radius  and  the  outer  radius 
of  which  are  respectively  rx  and  v% :  (1)  if  the  cylinder  is  homoge- 
neous ;  (2)  if  the  density  of  any  particle  is  proportional  to  its 
distance  from  the  axis  of  the  cylinder. 

96.  A  solid  is  bounded  by  the  plane  z  =  0,  the  cone  z  =  r  (cylindri- 
cal coordinates),  and  the  cylinder  having  its  elements  parallel  to  OZ 
and  its  directrix  one  loop  of  the  lemniscate  r  =  2  a2  cos  2  6  in  the 
plane  XOY.  Find  its  moment  of  inertia  about  OZ  if  the  density  at 
any  point  varies  directly  as  its  distance  from  OZ. 

97.  Find  the  moment  of  inertia  of  a  homogeneous  right  circular 
cone  of  density  p,  of  which  the  height  is  h  and  the  radius  of  the  base 
is  a,  about  an  axis  perpendicular  to  the  axis  of  the  cone  at  its  vertex. 

98.  A  ring  is  cut  from  a  homogeneous  spherical  shell  of  density  p, 
the  inner  radius  and  the  outer  radius  of  which  are  respectively  4  ft. 
and  5  ft.,  by  two  parallel  planes  on  the  same  side  of  the  center  of 
the  shell  and  distant  1  ft.  and  3  ft.  respectively  from  the  center. 
Find  the  moment  of  inertia  of  this  ring  about  its  axis. 

99.  A  mass  M  is  in  the  form  of  a  right  circular  cone  of  altitude 
h  and  with  a  vertical  angle  of  120°.  Find  its  moment  of  inertia 
about  its  axis  if  the  density  of  any  particle  is  proportional  to  its 
distance  from  the  base  of  the  cone. 

100.  The  radius  of  the  upper  base  and  the  radius  of  the  lower 
base  of  the  frustum  of  a  homogeneous  right  circular  cone  are  respec- 
tively &, 
its  axis. 


PEOBLEMS  401 

101.  The  density  of  any  point  of  a  solid  sphere  of  mass  M  and 
radius  a  is  directly  proportional  to  its  distance  from  a  diametral 
plane.  Find  its  moment  of  inertia  about  the  diameter  perpendicular 
to  the  above  diametral  plane. 

102.  Given  a  right  circular  cylinder  of  mass  M,  height  //,  and 
radius  a,  the  density  of  any  particle  of  which  is  /.•  times  its  distance 
from  the  lower  base.  Find  the  moment  of  inertia  of  this  cylinder 
about  a  diameter  of  its  lower  base. 

103.  Find  the  moment  of  inertia  about  (>Z  of  that  portion  of  the 
surface  of  the  hemisphere  z  =  V«"  —  x'~  —  f  which  lies  within  the 
cylinder  x"  +  if  =  ax. 

104.  A  homogeneous  solid  of  density  p  is  in  the  form  of  a  hemi- 
spherical shell,  the  inner  radius  and  the  outer  radius  of  which  are 
respectively  i\  and  /;,.  Find  its  moment  of  inertia  about  any  diam- 
eter of  the  base  of  the  shell. 

105.  A  homogeneous  anchor  ring  of  mass  .1/  is  bounded  by  the 
surface  generated  by  revolving  a  circle  of  radius  a  about  an  axis  in 
its  plane,  distant  b(b  >  a)  from  its  center.  Find  the  moment  of 
inertia  of  this  anchor  ring  about  its  axis. 

106.  The  density  at  any  point  of  the  hemisphere  z  =  "\Ar  —  x'2  —  >f 
is  k  times  its  distance  from  the  base  of  the  hemisphere.  Find  the 
moment  of  inertia  about  OZ  of  the  portion  of  the  hemisphere  in- 
cluded in  the  cylinder  x'2  +  //"  =  OX. 

107.  Through  a  homogeneous  spherical  shell  of  density  p,  of 
which  the  inner  radius  and  the  outer  radius  are  respectively  a^  and 
"„  a  circular  hole  of  radius  b(b<a^)  is  bored,  the  axis  of  the  hole 
coinciding  with  a  diameter  of  the  shell  Find  the  moment  of  inertia 
of  the  ring  thus  formed  about  the  axis  of  the  hole. 

108.  Find  the  center  of  gravity  of  the  portion  of  a  uniform  wire 
in  the  form  of  the  curve  x  =  at'2,  //  =  j-  <<f-\  ,-„■  =  \  af4}  between  the 
points  for  which  t  —  0  and  t  =  1. 

109.  Find  the  center  of  gravity  of  a  uniform  wire  in  the  form  of 
the  helix  x  =  a  cos  0,  y  =  a  sin  6,  z  =  k9,  between  the  points  for 
which  0  =  0  and  0  =  6V  "When  will  the  center  of  gravity  fall  on  the 
axis  of  the  helix  ? 

110.  Find  the  center  of  gravity  of  a  homogeneous  solid  bounded 
by  the  coordinate  planes  and  the  surface  x-  +  //-  +  ^  =  «5- 


402  MULTIPLE  INTEGRALS 

111.  Find  the  center  of  gravity  of  a  homogeneous  body  in  the 

form  of  an  octant  of  the  ellipsoid  '—  4-  7=  +  -z  =  1. 

1  a2       (r       c2 

112.  Eind  the  center  of  gravity  of  the  homogeneous  solid  bounded 
by  the  surfaces  z  —  0,  y  =  0,  y  =  b,  b2z2  =  y'2(a2  —  ar). 

113.  Eind  the  center  of  gravity  of  a  homogeneous  solid  bounded 
by  the  paraboloid  a2x2  +  b'2y2  =  z  and  the  plane  z  =  c. 

114.  A  ring  is  cut  from  a  homogeneous  spherical  shell  of  density 
p,  the  inner  radius  and  the  outer  radius  of  which  are  respectively 
4  ft.  and  5  ft.,  by  two  parallel  planes  on  the  same  side  of  the 
center  of  the  shell  and  distant  1  ft.  and  3  ft.  respectively  from  the 
center.    Eind  the  center  of  gravity  of  this  ring. 

115.  Eind  the  center  of  gravity  of  a  homogeneous  solid  bounded 
by  a  spherical  surface  of  radius  b  and  two  planes  passing  through 
its  center  and  including  a  dihedral  angle  2  a. 

116.  Find  the  center  of  gravity  of  a  hemisphere  of  radius  a  if 
the  density  at  any  point  varies  directly  as  the  distance  of  the  point 
from  the  base  of  the  hemisphere. 

117.  Find  the  center  of  gravity  of  a  homogeneous  solid  bounded 
by  the  surfaces  of  a  right  circular  cone  and  a  hemisphere  of  radius 
a  which  have  the  same  base  and  the  same  vertex. 

118.  Eind  the  center  of  gravity  of  an  octant  of  a  sphere  of  radius 
a  if  the  density  at  any  point  varies  directly  as  its  distance  from  the 
center  of  the  sphere. 

119.  Find  the  center  of  gravity  of  a  right  circular  cone  of  altitude 
a,  the  density  of  each  circular  slice  of  which  varies  directly  as  the 
square  of  its  distance  from  the  vertex. 

120.  Eind  the  center  of  gravity  of  a  homogeneous  solid  bounded 
by  two  concentric  spherical  surfaces  of  radii  4  ft.  and  5  ft.  respec- 
tively and  a  plane  through  the  common  center  of  the  two  spherical 
surfaces. 

121.  Find  the  center  of  gravity  of  a  homogeneous  solid  in  the 
form  of  the  frustum  of  a  right  circular  cone,  the  height  of  which 
is  h  and  the  radius  of  the  upper  base  and  the  radius  of  the  lower 
base  of  which  are  respectively  r%  and  rr 


PROBLEMS  403 

122.  A  solid  is  bounded  by  a  sphere  of  radius  a  and  a  right  circu- 

7T 

lar  cone,  the  vertical  angle  of  which  is  —  >  the  vertex  of  which  is  on 

o 

the  surface  of  the  sphere,  and  the  axis  of  which  coincides  with  a 

diameter  of  the  sphere.    Find  its  center  of  gravity  if  the  density  at 

any  point  is  k  times  its  distance  from  the  axis  of  the  cone. 

123.  Find  the  attraction  of  a  hemisphere  of  radius  a  on  a  particle 
of  unit  mass  at  the  center  of  its  base  if  the  density  at  any  point  of 
the  hemisphere  varies  directly  as  its  distance  from  the  base. 

124.  A  homogeneous  solid  of  density  p  is  bounded  by  the  plane 
z  =  3  and  the  surface  z%  ==  x1  +  if.  Find  the  attraction  of  this  solid 
on  a  particle  of  unit  mass  at  the  origin  of  coordinates. 

125.  A  portion  of  a  right  circular  cylinder  of  radius  <i  and  uniform 
density  p  is  bounded  by  a  spherical  surface  of  radius  b(b>a),  the 
center  of  which  coincides  with  the  center  of  the  base  of  the  cylinder. 
Find  the  attraction  of  this  portion  of  the  cylinder  on  a  particle  of 
unit  mass  at  the  middle  point  of  its' base 

126.  A  portion  of  a  right  circular  cylinder  of  radius  a  is  bounded 
by  a  spherical  surface  of  radius  b(b>a),  the  center  of  winch  coin- 
cides with  the  center  of  the  base  of  the  cylinder.  Find  the  attraction 
of  this  portion  of  the  cylinder  on  a  particle  of  unit  mass  at  the 
middle  point  of  its  base,  the  density  of  any  particle  of  the  cylinder 
being  proportional  to  its  distance  from  the  axis  of  the  cylinder. 

127.  Show  that  the  attraction  of  a  segment  of  one  base,  cut  from 
a  homogeneous  sphere  of  radius  a,  on  a  particle  of  unit  mass  at  its 

vertex  is  2ir/ip(l  —  -  ^l_),  where  p  is  the  density  of  the  sphere 

and  h  is  the  height  of  the  segment. 

128.  A  ring  is  cut  from  a  homogeneous  spherical  shell  of  density 
p,  the  inner  radius  and  the  outer  radius  of  which  are  respectively 
4  ft.  and  5  ft.,  by  two  parallel  planes  on  the  same  side  of  the  center 
of  the  shell  and  distant  1  ft.  and  3  ft.  respectively  from  the  center. 
Find  the  attraction  of  this  ring  on  a  particle  of  unit  mass  at  the 
center  of  the  shell. 

129.  The  density  of  a  hemisphere  of  mass  ^^  and  radius  a  varies 
directly  as  the  distance  from  the  base.  Find  its  attraction  on  a 
particle  of  unit  mass  in  the  straight  line  perpendicular  to  the  base 


404  MULTIPLE  INTEGRALS 

at  its  center,  and  at  a  distance  a  from  the  base  in  the  direction 
away  from  the  hemisphere. 

130.  A  solid  of  mass  M  is  bounded  by  a  right  circular  cone  of 
vertical  angle  90°  and  a  spherical  surface  of  radius  2  ft.,  the  center 
of  the  spherical  surface  being  at  the  vertex  of  the  cone.  If  the 
density  of  any  particle  of  the  above  solid  is  directly  proportional  to 
its  distance  from  the  vertex  of  the  cone,  find  the  attraction  of  the 
solid  on  a  particle  of  unit  mass  at  the  vertex  of  the  cone. 

131.  The  vertex  of  a  right  circular  cone  of  vertical  angle  2  a  is 
at  the  center  of  a  homogeneous  spherical  shell,  the  inner  radius  and 
the  outer  radius  of  which  are  respectively  ax  and  «2.  Eind  the  attrac- 
tion of  the  portion  of  the  shell  outside  the  cone  on  a  particle  of  unit 
mass  at  the  center  of  the  shell,  in  terms  of  the  attracting  mass. 

132.  The  density  at  any  point  of  a  given  solid  of  mass  M  in  the 
form  of  a  hollow  right  circular  cylinder  is  directly  proportional  to 
its  distance  from  the  axis  of  the  cylinder.  If  the  height  of  the 
cylinder  is  2  ft.,  and  its  inner  radius  and  outer  radius  are  respec- 
tively 1  ft.  and  2  ft.,  find  its  attraction  on  a  particle  of  unit  mass 
situated  on  its  axis  2  ft.  below  the  base. 


CHAPTER  XVII 
INFINITE  SERIES 

187.  Convergence.    The  expression 

«l  +  a1  +  aI"+a4  +  aB  +  ---,  (1) 

where  the  number  of  the  terms  is  unlimited,  is  called  an  infinite 
semes. 

An  infinite  series  is  said  to  converge,  or  to  be  convergent,  when  tin- 
sum  of  the  first  n  terms  ajyproaches  a  limit  as  n  increases  without  limit. 

Thus,  referring  to  (1),  we  may  place 

s„  =a1+a2, 


Then,  if  Lim8n=  A, 

the  series  is  said  to  converge  to  the  limit  ./.  The  quantity  A  is 
frequently  called  the  sum  of  the  series,  although,  strict  ly  speaking, 
it  is  the  limit  of  the  sum  of  the  first  n  terms.  The  convergence 
of  (1)  may  be  seen  graphically  by  plotting  *r  *.,,  *..,  •  •  •,  8n  on  the 
number  scale,  as  in  §  3. 

A  series  which  is  not  convergent  is  called  divergent.  This  may 
happen  in  two  ways :  either  the  sum  of  the  first  n  terms  increases 
without  limit  as  n  increases  without  limit ;  or  sn  may  fail  to  approach 
a  limit,  but  without  becoming  indefinitely  great. 

Ex.  1.    Consider  the  geometric  series 

a  +  ar  +  ar"  +  <irz  +  •  •  • . 

1  —  r" 

Here  sn  =  a  +  ar  +  ar2  +  •  •  •  +  ar"-1  =  a- Xow  if  r  is  numerically 

\  —  r 

less  than  1,  r"  approaches  zero  as  a  limit  as  n  increases  without  limit ;  and 

ac  405 


406  INFINITE  SERIES 

therefore  Lim  sn  = If,  however,  r  is  numerically  greater  than  1, 

rn  increases  without  limit  as  n  increases  without  limit ;  and  therefore  sn 
increases  without  limit.    If  r  =  1,  the  series  is 

and  therefore  sn  increases  without  limit  with  n.    If  r  =  —  1,  the  series  is 

a  —  a  +  a  —  «+•••, 
and  sn  is  alternately  a  and  0,  and  hence  does  not  approach  a  limit. 

Therefore,  the  geometric  series  converges  to  the  limit when  r  is  numeri- 
cally less  than  unity,  and  diverges  when  r  is  numerically  equal  to,  or  greater 
than,  unity. 

Ex.  2.    Consider  the  harmonic  series 

1  +  I  +  I  +  1  +  1+1  +  I  +  I  +  ...  +  1  +  .;., 

2345078  n 

consisting  of  the  sum  of  the  reciprocals  of  the  positive  integers.    Now 

£  +  £  >  \  +  k  =  h      HHI  +  ?>H1+Hi  =  l. 

and  in  this  way  the  sum  of  the  first  n  terms  of  the  series  may  be  seen  to 
be  greater  than  any  multiple  of  ^  for  a  sufficiently  large  n.  Hence  the 
harmonic  series  diverges. 

188.  The  comparison  test  for  convergence.  If  each  term  of  a 
given  series  of  positive  numbers  is  less  than,  or  equal  to,  the  corre- 
sponding term  of  a  known  convergent  series,  the  given  series 
converges. 

If  each  term  of  a  given  series  is  greater  than,  or  equal  to,  the 
corresponding  term  of  a  known  divergent  series  of  positive  numbers, 
the  given  series  diverges. 

Let  «1+«2+rt3+fl,4+ •••  C1) 

be  a  given  series  in  which  each  term  is  a  positive  number,  and  let 

K+h+h+\+---  (2) 

be  a  known  convergent  series  such  that  ak  ^  bk. 

Then,  if  sn  is  the  sum  of  the  first  n  terms  of  (1),  s'n  the  sum 
of  the  first  n  terms  of  (2),  and  B  the  limit  of  s'n,  it  follows  that 


CONVEEGENCE  407 

since  all  terms  of  (1),  and  therefore  of  (2),  are  positive.  Now  as 
n  increases,  sn  increases  but  always  remains  less  than  B.  Hence  sH 
approaches  a  limit,  which  is  either  less  than,  or  equal  to,  B. 

The  first  part  of  the  theorem  is  now  proved ;  the  second  part 
is  too  obvious  to  need  formal  proof. 

In  applying  this  test  it  is  not  necessary  to  begin  with  the 
first  term  of  either  series,  but  with  any  convenient  term.  The 
terms  before  those  with  which  comparison  begins  form  a  poly- 
nomial, the  value  of  which  is  of  course  finite,  and  the  remaining 
terms  form  the  infinite  series  the  convergence  of  which  is  to  be 
determined. 


+  -,  +  -  +  -+•■•  + 


Ex.  1.    Consider 

1      2s      :P      i*  (n-1)" 

Each  term  after  the  third  is  less  than  the  corresponding  term  of  the  con- 
vergent geometric  .scries 

i  +  i+^  +  i  +  i+...  +  5li  +  .... 

Therefore  the  first  series  converges. 

Ex.  2.    Consider  % 

V2       V3      Vi       V5  VTi 

Each  term  after  the  first  is  greater  than  the  corresponding  term  of  the 

divergent  harmonic  scries 

Therefore  the  first  series  diverges. 

189.  The  ratio  test  for  convergence.  If  in  a  series  of  positive 
numbers  the  ratio  of  the  (>i+l)s.  term  to  the  nth  term  approaches 
a  limit  L  as  n  increases  without  limit,  then,  if  L<\,  the  series 
converges ;  if  L>1,  the  series  diverges ;  if  L  —  1,  the  series  may 
either  diverge  or  converge. 

Let  a,  +  aa  +  a3  -\ +  an  +  an+1  H (1) 

be  a  series  of  positive  numbers,  and  let  Limi-i±i=X.    We  have 
three  cases  to  consider.  "=:°      " 


408  INFINITE  SERIES 

1.  L<1.  Take  r  any  number  such  that  i<r<l.  Then, 
since  the  ratio  -Ail  approaches  L  as  a  limit,  this  ratio  must 

become  and  remain  less  than  r  for  sufficiently  large  values  of  n. 
Let  the  ratio  be  less  than  r  for  the  rath  and  all  subsequent 
terms.    Then  am  +  1<aiur, 

am+-2<am  +  ir<a„A 
Now  compare  the  series 

am  +  ^m  +  l  +  «m  +  2  +  «m  +  3H (2) 

with  the  series  am  +  amr  +  amr2  -f  amr3  +  •  •  •.  (3) 

Each  term  of  (2)  except  the  first  is  less  than  the  correspond- 
ing term  of  (3),  and  (3)  is  a  convergent  series  since  it  is  a 
geometric  series  with  its  ratio  less  than  unity.  Hence  (2) 
converges  by  the  comparison  test,  and  therefore  (1)  converges. 

2.  L>1.    Since  -JL±1  approaches  I  as  a  limit  as  n  increases 

without  limit,  this  ratio  eventually  becomes  and  remains  greater 
than  unity.  Suppose  this  happens  for  the  wth  and  all  subsequent 
terms.    Then  ^ 

am+\  ~^am1 

«m  +  2>am+l>V 


Each  term  of  the  series  (2)  is  greater  than  the  corresponding 
term  of  the  divergent  series 

a»  +  «»  +  ««  +  fflJ •  (4) 

Hence  (2)  and  therefore  (1)  diverges. 

3.  L  =1.  Neither  of  the  preceding  arguments  is  valid,  and 
examples  show  that  in  this  case  the  series  may  either  converge 
or  diverge. 

In  applying  this  test,  the  student  will  usually  find  -R±1  hi 

the  form  of  a  fraction  involving  n.     To  find  the  limit  of  this 


CONVERGENCE 


409 


fraction   as  n  increases  without  limit,   it  is   often  possible  to 
divide  numerator  and  denominator  by  some  power  of  n,  so  as 

to  be  able  to  apply  the  theorem  (§13)  that  Lim-  =  0,  or  some 
other  known  theorem  of  limits. 

Ex.  1.    Consider 

2       3       4       5  n 


3»- 


The  nth  term  is  — — -  and  the  (n  +  l)st  term  is 

(n  +  l)st  term  to  the  nth  term  is ,  and 

3ft  j 

T  .      n  + 1       T  .  n      1 

Lim =  Lim =  - 

n=»    3fl         „=„      3  3 


+  1 


The  ratio  of  the 


Therefore  the  given  series  converges. 
Ex.  2.   Consider     1  + 


38      44 


+  r  + 


The  nth  term  is  — ■  and  the  (n  +  l)st  term  is 


(»  +  l)'- 


the  (n  +  l)st  term  to  the  nth  term  is 


(n +  !)»  +  ' 


n  +  1 

-  =  £+*)" 
(n  +  L)»"       \    n    / 

Lto(!i±l)-=Lim(l  +  l) 

n  =  »  \     ft     /        n  =  ao  \         n/ 
Therefore  the  given  series  diverges. 


The  ratio  of 


,  and 


(§98) 


190.  Absolute  convergence.  The  absolute  value  of  a  real  num- 
ber is  its  arithmetical  value  independent  of  its  algebraic  sign. 
Thus  the  absolute  value  of  both  +  2  and  —  2  is  2.  The  abso- 
lute value  of  a  quantity  a  is  often  indicated  by  \a\.  It  is  evi- 
dent that  the  absolute  value  of  the  sum  of  n  quantities  is 
less  than,  or  equal  to,  the  sum  of  the  absolute  values  of 
the  quantities. 

A  series  converges  when  the  absolute  values  of  its  terms  form  a 
convergent  series,  and  is  said  to  converge  absolutely. 

Let 


ai  +  a2  +  az  +  ai~\ 

•be  a  given  series,  and 


(i) 

(2) 


410  INFINITE  SERIES 

the  series  formed  by  replacing  each  term  of  (1)  by  its  absolute 
value.  We  assume  that  (2)  converges,  and  wish  to  show  the 
convergence  of  (1). 

Form  the  auxiliary  series 

(«i  +  KI)  +  («k  +  N)  +  («.  +  |«,|)  +  (a4  +  |aJ)+....     (3) 

The  terms  of  (3)  are  either  zero  or  twice  the  corresponding 
terms  of  (2).  For  ak  =  —  \ak\  when  ak  is  negative,  and  ak  =  \ak\ 
when  ak  is  positive. 

Now,  by  hypothesis,  (2)  converges,  and  hence  the  series 

2W+2|«.l+2|«1|+2|aJ+...  (4) 

converges.  But  each  term  of  (3)  is  either  equal  to  or  less 
than  the  corresponding  term  of  (4),  and  hence  (3)  converges  by 
the  comparison  test. 

Now  let  sn  be  the  sum  of  the  first  n  terms  of  (1),  s'n  the  sum 
of  the  first  n  terms  of  (2),  and  s'J  the. sum  of  the  first  n  terms 
of  (3).    Then  '    sn  =  s'J-s'n, 

and,  since  s"  and  s'n  approach  limits,  sn  also  approaches  a  limit. 
Hence  the  series  (1)  converges. 

We  shall  consider  in  this  chapter  only  absolute  convergence. 
Hence  the  tests  of  §§  188,  189  may  be  applied,  since  in  testhig 
for  absolute  convergence  all  terms  are  considered  positive. 

191.  The  power  series.    A  power  series  is  defined  by 

%  +  v;  +  v;2  +  azx*  ^ •■  anx"  -\ —  > 

where  «0,  a ,  a0,  a8,  ■  •  ■  are  numbers  not  involving  x. 

We  shall  prove  the  following  theorem :  If  a  'power  series  con- 
verges for  x  =  xv  it  converges  absolutely  for  any  value  of  x  such 
that  \x\<\xx\. 

For  convenience,  let  \x\  =  X,  \an\  =  An,  jzj  =  A'a.  By  hyjDothe- 
sis  the  series 

a0  +  axxx  +  a„x\  +  azx\  -\ h  anx[l  H (1) 

converges,  and  we  wish  to  show  that 

A0  +  AXX+  AX'  +  AZX*  +  •  •  •  +  AnX'1  +  •  •  •  (2) 

converges  if  X<X,. 


POWER  SERIES  411 

Since  (1)  converges,  all  its  terms  are  finite.  Consequently 
there  must  be  numbers  winch  are  greater  than  the  absolute 
value  of  any  term  of  (1).  Let  M  be  one  such  number.  Then 
we  have  AnX"  <  M  for  all  values  of  n. 

Then  /  TV  IX\* 

Each  term  of  the  series  (2)  is  therefore  less  than  the  corre- 
sponding term  of  the  series 

M+  M(f)+M(§)'+  u(§]+  •  ■  • +  *©"+  "  -  (3) 

But  (3)  is  a  geometric  series,  which  converges  when  X<  Xx. 
Hence,  by  the  comparison  test,  (2)  converges  when  X<X. 

From  the  preceding  discussion  it  follows  that  a  power  series 
will  converge  for  values  of  x  lying  between  two  numbers  —  B 
and  +  B,  and  diverge  for  all  other  values  of  x.  In  some  cases 
B  may  be  infinity,  that  is,  the  series  may  converge  for  all  values 
of  x.  In  other  cases,  less  frequent,  B  may  be  zero,  that  is,  the 
series  may  converge  only  for  x  =  0. 

In  any  case  the  values  of  x  for  which  the  series  converges 
are  together  called  the  region  of  convergence.  If  represented 
on  a  number  scale,  the  region  of  convergence  is  in  general  a 
portion  of  the  scale  having  the  zero  point  as  its  middle  point. 
In  some  cases  the  region  may  extend  to  infinity  or  shrink  to 
a  point.  In  practice  the  student  will  generally  find  it  con- 
venient to  determine  the  region  of  convergence  by  applying 
the  ratio  test,    as  shown  hi  the  examples. 

Ex.  1.   Consider 

1  +  2  x  +  3  x2  +  4  xs  +  •  •  •  +  nx"-1  +  ■  ■  -. 

The  nth  term  is  nxn-1,  the  (n  +  l)st  term  is  (n  +  l)xn,  and  their  ratio  is 

x.   Lim x  —  x  Lim  (1  +  -)  =  x.   Hence  the  series  converges  when 

n  „  =  *,      n  n  =  oo\         n) 

\x\  <1  and  diverges  when  \x\  >1.    The  region  of  convergence  extends  on 

the  number  scale  between  —  1  and  +  1. 


412  INFINITE  SERIES 

Ex.  2.    Consider  1  + Z +  £. +  ±  +  . . .  +  JL_ -  +  . . .. 
1      [2      13  | ».  —  1 

— -j  the  (n  +  l)st  term  is  — ,  and  their  ratio  is  -• 

|n  —  1  [n  ra 

Liin  -  =  0  for  any  finite  value  of  x.  Hence  the  series  converges  for  any 
value  of  x  and  its  region  of  convergence  covers  the  entire  number  scale. 

Ex.  3.   Consider 

1  +  x  +  (2  x2  +  [3  Xs  +  •  •  •  +  Ira-la;"-1  +  •  •  •. 

The  nth  term  is  \n  —  la;"-1,  the  (n  +  l)st  term  is  \nxn,  and  their  ratio 
is  nx.  This  ratio  increases  without  limit  for  all  values  of  x  except  x  =  0. 
Therefore  the  series  converges  for  no  value  of  x  except  x  —  0. 

A  power  series  defines  a  function  of  x  for  values  of  x  within  the 
region  of  convergence,  and  we  may  write 

f(x)  =%+  axx  +  a^x"  +  a3x3  -\ +  a„of  +  •  •  • ,  (4) 

it  being  understood  that  the  value  of  f(x)  is  the  limit  of  the 
sum  of  the  series  on  the  right  of  the  equation.  The  power 
series  has  the  important  property,  not  possessed  by  all  kinds 
of  series,  of  behaving  very  similarly  to  a  polynomial.  When  a 
function  is  expressed  as  a  power  series  it  may  be  integrated 
or  differentiated  by  integrating  or  differentiating  the  series  term 
by  term.  The  new  series  will  be  valid  for  the  same  values  of 
the  variable  for  which  the  original  series  is  valid.  If  the  method 
is  applied  to  a  definite  integral,  the  limits  must  be  values  for 
which  the  series  is  valid. 

Similarly,  if  two  functions  are  each  expressed  by  a  power  series, 
their  sum,  difference,  product,  or  quotient  is  the  sum,  the  dif- 
ference, the  product,  or  the  quotient  of  the  series. 

For  proofs  of  these  theorems  the  student  is  referred  to  ad- 
vanced treatises. 

192.  Maclaurin's  and  Taylor's  series.  We  have  noted  that  any 
convergent  power  series  may  define  a  function.  Conversely, 
it  may  be  shown  (see  §  193)  that  any  function  which  is  con- 
tinuous and  has  continuous  derivatives  may  be  expressed  as  a 
power  series.    When  a  function  is  so  expressed  it  is  possible  to 


«o=/(°)'  «x=/(0).  V=|/'(°)>  «3=|/"'(°)'  •  •  ">  ^=^/(n)(0). 


MACLAUEIX'S  AND  TAYLOE'S  SEEIES  413 

express  the  coefficients  of  the  series  in  terms  of  the  f miction 

and  its  derivatives.    For  let 

f(x)=  %+  arv  +  ajr+  a3x3  +  «/+•••+  a„af + 

By  differentiating  we  have 

f(x)=a1+  2  a2x  +  3  «,**  +  4  a4x3H +  na^'1-] , 

f"(x)=  2  «2+  3  •  2  rt8.r  +  4  •  3  «/+  •  •  •  +  n(n-l)«yf-f+  •  •  •, 

f"(x)  =  S.  2a8+4  •  3  •  2a4«+  •  •  ■  +»(n-l)(n-2)a.af-»+  ■  •  ■, 

/(%r)  =  [>(W  -1)(h-  2)  •  •  •  3  •  2]an  +  ■  •  -. 

Placing  £=0  in  each  of  these  equations,  we  find 
=/(0),  a=f(0-),a  = 

Consequently  we  have 

/(.O=/(0)+/X0),+-^..+^,-+...+-^.+  ..,(i) 

This  is  called  Maclaurin,8  series. 
Again,  if  in  the  right-hand  side  of 

f(x)  =  aQ  +  atx  +  ./.,/•-  +  ay  +  . . .  +  ,/„.>•"  +  . . . 
we  place  x  =  a  4-  •>•',  and  arrange  according  to  powers  of  a/,  we  have 

f(x)  =  ?>o  +  6/  +  6aa/a  +  ft/8  +  •  •  •  +  ?>,/»  +  . . ., 
or,  by  replacing  ./•'  by  its  value  a;  — a, 
/(.r)  =  60+  ^(s - a)+  \(x  -  a)a+  6,(a: -  a ):,+  . .  •  +  K(x-  a)*+ 

By  differentiating  this  equation  successively,  and  placing  x=a 
in  the  results,  we  readily  find 

».=/oo.  Woo.  &a=,4/'oo,  ».-[l/"(«).-.  j-=^/(b)(«)- 

Hence 

/w-««)+c»-«)f«+^/'(«)+^Vw+  •  •  • 

+-*^^.r00+----  (2) 

This  is  Taylors  series. 


414  INFINITE  SERIES 

Another  convenient  form  of  (2)  is  obtained  by  placing 
x  —  a  =  h,  whence  x  =  a  +  h.    We  have  then 

/(«  +  h)  =/(«)  +  hf(a)  +~f\a)  +^f"(a)  +  .  ■  ■ 
If  If 

Maclaurin's  series  (1)  enables  its  to  expand  a  function  into 
a  series  in  terms  of  ascending  powers  of  x  when  we  know  the 
value  of  the  function  and  its  derivatives  for  x  =  0.  By  means 
of  the  series  the  function  may  be  computed  for  values  of  x  for 
which  the  series  converges.  Practically  the  computation  is  con- 
venient for  small  values  of  x. 

Taylor's  series  (2)  enables  us  to  expand  a  function  in  terms 
of  powers  of  x  —  a  when  the  value  of  the  function  and  its 
derivatives  are  known  for  x  =  a.  The  function  is  said  to  be 
expanded  in  the  neighborhood  of  x  =  a,  and  the  series  can  be 
used  to  compute  the  value  of  the  function  for  values  of  x 
which  are  near  a. 

Ex.  1.    Expand  ex  into  a  power  series  and  compute  its  value  when  x  =  \. 
Since  f(x)  =  e*,  f  (x)  =  e*,  f"  (x)  =  eF,  etc.,  /(0)  =  1,  f  (0)  =  1,  f"  (0)  =  1, 
etc.    Hence,  by  Maclaurin's.  series, 

x      .r2      x3      xi 
e*  =  1  +  l+j.+ii  +  j4  +  -"- 

This  converges  for  all  values  of  x.  If  we  place  x  =  J,  we  have  e*  =  1 
+  i  +  is  +  t<?2  +  tsjW  =  1-3956,  correct  to  four  decimal  places.  If  x  has 
a  larger  value,  more  terms  of  the  series  must  be  taken  in  the  computation, 
so  that  the  series,  while  valid,  is  inconvenient  for  large  values  of  x. 

Ex.  2.    Expand  (a  +  a;)"  into  a  power  series  in  x.    Here 
f{x)  =  (n  +  xy,  /(0)  =  a», 

f(x)  =  n(a  +  x)n~1,  f'(0)  =  nan-1, 

f"  (x)  =  n  (n  - 1)  (a  +  x)«  ~  2,  f"  (0)  =  n  (n  - 1)  a»  -  2, 

f"  (x)  =  n  (n  -  1)  (n  -  2)  (a  +  x)n  -  3,    f"  (0)  =  n  (n  - 1)  (n  -  2)  a"'3. 

Hence,  by  Maclaurin's  series, 

(a  +  xy  =  a"  +  na»-ix  +  n(n~l)  (l"-*x2  +  "  ("  ~  ^  (n  ~  2>>  a«~3xs  +  •  ■  -. 


MACLAURIN'S  AND  TAYLOE'S  SERIES  415 

This  is  the  binomial  theorem.  If  n  is  a  positive  integer,  the  expansion  is 
a  polynomial  of  n  +  1  terms,  since  f(n  +  1)(x)  and  all  higher  derivatives  are 
equal  to  0.  But  if  n  is  a  negative  integer  or  a  fraction,  the  series  converges 
when  x  is  numerically  less  than  a. 

Ex.  3.    Find  the  value  of  sin  61°. 

Let  f(x)  =  sinx,  then  f'(x)  =  cosx,  f  (x)  =  —  sinx,  etc.,  provided  x  is 

expressed  in  circular  measure.    61°  expressed  in  circular  measure  is  ^Vo  7r 

=  —  +  -^— .    Since   sin  -   and   cos  -   are    known   to   be   respectively   -  v3 
3      180  3  3  *  J    2 

and  — ,  it  will  be  convenient  to  use  Taylor's  series  with  a  =  -  ■    Formula 
2  J  3 

(it    ,     ,\  .      I,,  7T         h-     .      IT         ft8  7T 

-  +  h    =  sin  -  +  h  cos  -  —  —  sin  -  —  —  cos  -  +  •  •  • 
^3        /  3  3      [2        3      [3        a 


ivi+J»-^-ip+ 


Placing  /*  =  — —  and  computing,  we  have 


^(ll*) 


sin  61°  =  .S74G. 


The  expansion  of  a  function  may  sometimes  be  obtained  by- 
special  devices  more  conveniently  than  by  direct  use  of  the  for- 
mula (1)  or  (2):   This  is  illustrated  by  the  following  examples  : 

Ex.  4.    Required  to  expand  sin_1x. 
We  have 


sin-ix=  fX  =  fX(l-x*)-ldx 

Jo  VT^~72     J° 


Io\1  +  h2  +  hixi  +  hT^x6  +  --)'lx  (1,yEx-2) 


1   x*     1-3    x5      1-3-5    x7 
~X  +  2'32-4'52-4-6'7 


„      .     m  i    sm_1x 

Ex.  5.    To  expand 


Vl  -  x2 

x3      3  x5      5  x7 
ByEx.4,  sin-ix  =  x  +  |  +  i£.  +  ^  +  ..- 

i  r2      3  x4      5  x6 

by  Ex.2,  (l_,Tl=l  +  |.  +  ^_ +  !_+... 

Hence,  by  multiplication, 

sin-1  j  2x3      Sx5      16  x7 

Vf^~c*~  3         15         35 


416  INFINITE  SERIES 

193.  The  remainder  in  Taylor's  series.    Let  us  write 

/(*)  =/(«)  +  O  -  «■)/'  00  +  ^w^-f"  00  +  •  •  • 

and  attempt  to  determine  R.    For  that  purpose  place 

In  the  right-hand  member  of  equation  (1),  with  P  in  the 
form  (2),  replace  a  everywhere,  except  in  P,  by  z,  and  call 
F(z)  the  difference  between  f(x)  and  this  new  expression. 
That  is,  let  F(z)  be  defined  by  the  equation 

f(?)  =f(p)  -m  -  o  -  *)/'  oo  -  (J~^r  oo  -  ■  ■  • 

5~y  (*}  — khT  p'  (3) 

where  x  is  considered  constant. 

Differentiate  (3)  with  respect  to  z,  still  holding  z  constant. 
All  the  terms  obtained  cancel,  except  the  last  two,  and  we  have 

f>  <» = -  ^f^> +i)  (z) + (*^y  P 

=^£):[P_/(re+1)(g)]-  (4) 

IZ: 

Now  when  2  —  2*,  P(2)=0,  as  is  at  once  apparent  from  (3). 
Also  when  z  —  a,  F(^z)  =  0,  as  appears  from  (3)  with  the  aid 
of  (1).  Hence  F(z)  must  have  a  maximum  or  a  minimum  for 
some  (unknown)  value  of  z  between  z  =  a  and  z  —  x.    That  is, 

F'Q)  =  0 
where  f  lies  between  a  and  x.* 

*  The  theorem  that  if  F(z)  =  Ofor  z  =  a  and  z  =  b,  then  F'(z)  =  0  for  some 
value  of  z  between  z  —  a  and  z  =  b  is  called  Rolle's  Theorem.  It  is  geometrically 
evident  on  drawing  a  graph.  Of  course  F(z)  and  F'(z)  must  be  continuous  and 
hence  finite. 


REMAINDER  IX  TAYLOR'S  SERIES  417 

From  (4),  it  follows  that 

p=/(b+1)(£>, 

and  hence,  from  (2), 

S  =  ^fff^(?)-  (5) 

This  is  the  remainder  in  Taylor  s  Theorem.  It  measures  the 
difference  between  the  value  of  the  function /(./)  and  the  sum 
of  the  first  n  +  1  terms  in  (1). 

It  is  evident  that  if  R  approaches  zero  as  n  is  indefinitely 
increased,  the  Taylor's  series  converges  and  represents  the 
function.  We  have,  then,  in  this  case,  a  proof  of  the  possi- 
bility of  a  series  expansion  for  the  function,  which  was  assumed 
in  §  192. 

Generally  also  it  will  be  sufficient  to  test  the  convergence 
of  the  series  by  one  of  the  methods  of  §§  188  and  189.  For 
usually  if  the  series  converges,  it  properly  represents  the  func- 
tion. Examples  can  be  given  in  which  this  is  not  true,  but  the 
student  will  certainly  not  meet  them  in  practice. 

The  remainder  may  be  said  to  measure  the  error  made  in  cal- 
culating the  value  oif(x)  by  means  of  n  +  1  terms  of  a  Taylor's 
or   Maelaurin's  series.     It  is  therefore  often   important    to  know 

Bometihing  of  the  magnitude  of  R.   Now  H  can  usually  nut  be 

found  exactly,  since  £  is  unknown,  but  it  can  sometimes  be  seen 
that  B  cannot  exceed  some  known  value,  and  this  is  enough  for 
practice.     This  is  illustrated  in  the  examples. 

Ex.  1.  What  error  is  made  by  calculating  <  8  by  5  terms  of  Maelaurin's 
series?    (See  Ex.1,  §  102.) 

When/(x)=  e?r,/<B+1>(a:)  =  <■.    Hence,  in  Maelaurin's  series  for  e*, 
n  +  ] 


where  £  lies  between  0  and  x. 

In  the  present  example  n  =  4  and  x  =  3. 

Therefore  11  =^e^  =  — —  & 

5  29160 


418  INFINITE  SERIES 

where  £  lies  between  0  and  J.    Since  the  largest  value  of  £  gives  the 
largest  value  of  e$,  we  may  write 

-ft  <  23  i  so e3  <  ^^  f  <nr 3? ; 

whence  it  appears  that  .R<. 00005. 

The  calculation  of  Ex.  1,  §  192,  is  therefore  correct  to  4  decimal  places. 

Ex.  2.    How  many  terms  of  Maclaurin's  series  must  be  taken  to  compute 

e?  correctly  to  4  decimal  places  ? 

CAY1*1   * 
As  in  Ex.1,  2J=I*2_^ 

\n  +  1 


where  $  is  between  0  and  \.   Hence 


h  +  1 


and  n  +  1  must  be  so  determined  that 


£i^32<.00005. 
n  +  1 


This  can  be  done  only  by  trial.  It  results  that  n  +  1  =  6.  Then  6  terms 
will  be  sufficient  to  assure  the  required  accuracy,  though  from  the  nature 
of  the  calculation  fewer  terms  may  do. 

194.  Relations  between  the  exponential  and  the  trigonometric 
functions.    By  Maclaurin's  series,  we  find 

^i+i+i+i+T(i+--->        (1) 

x3      x5      x1 
sin*  =  ;c__  +  ___+...>  (2) 

cos,  =  l-|  +  |-^+..,  (3) 

where  the  laws  governing  the  terms  are  evident.  It  is  possible 
to  show  that  in  each  case  R  approaches  zero  as  the  number  of 
the  terms  increases  without  limit,  no  matter  what  the  value  of  x. 
Hence  the  series  converge  and  represent  the  functions  for  all 
real  values  of  x. 

The  series  (1)  may  be  used  to  define  the  meaning  of  e*  when 
x  is  a  pure  imaginary  quantity  and  the  definitions  of  §  26  no 


APPROXIMATE  INTEGRATION  419 

longer  have  a  meaning.    We  write  as  usual  i  =  V—  1  and  replace 
x  in  (1)  by  ix.    We  obtain 

-i  +  T  +  "]2~  +  "]3_  +  "]^+•"• 
Then,  since     i*  =  —  l,         i3=  —  i,         &*=+l,  etc., 

fa     A      *2      *4  \,    ■/        *3  ,  *5  \ 

•-f-i+g— H*;ii+s— ■> 

But  the  two  series  here  involved  are  equal  to  cos  a;  and  sin  a; 
respectively  by  (3)  and  (2).    Hence  we  have 

eix  =  cos  x  +  i  sin  x.  (4) 

Similarly,  e~ ix  =  cos  #  —  i  sin  #,  (5) 

and,  from  (4)  and  (5), 

(6) 

(7) 

The  results  (4)-(7)  are  of  great  importance  in  some  appli- 
cations, notably  to  the  simplification  of  certain  results  in  the 
solution  of  differential  equations. 

It  may  be  proved  from  (1)  that  t?leXt  =  eXl+x-.  Then 

e ~ iy  =  (fe- ilJ  =  ex  (cos  y  -  i  sil i  y).  (9) 

195.  Approximate  integration.  When  it  is  not  possible,  or 
convenient,  to  evaluate  the  integral 


sin  a; 

= 

>■'■ 

—  e 

2i 

cos  # 

e* 

+  e~ 

ix 

I. 


f(x)dx  (1) 


exactly,  the  function  f(x)  may  be  expanded  into  a  power  series 
and  the  integral  computed  to  any  required  degree  of  accuracy. 
This  procedure  leads  to  the  following  three  rules: 

1.  TJie  prismoidal  formula.  Let  us  take  the  first  four  terms 
of  Taylor's  series  for  f(x)  in  the  neighborhood  of  x  =  «,  writ- 
ing them  hi  the  form 

f(x)  =A  +  B  (x  -a)+C  (x  -  a)2  +  D  (x  -  a)9.  (2) 


420  INFINITE  SEEIES 

Substituting  this  in  (1),  we  have 


I 


f(x)dx  =  A(b-a}+lBQ)-ay  +  ±C(b-ay  +  \D(l>-ay 

=  ^[6^+3jB(5-a)+2C(6-a)2+fZ>(6-a)8].  (3) 


Now,  from  (2), 

/(«)  =  4 

/(J)  =  .4  +  /?  (J  _  a)  +  C(J  -  ay  +  D(b-  a)3, 

and      /  fe±±\  =  l  +  }5(i-«)+j  C(J  -  a)2  +  |  Z>(5  -  a)3 ; 

from  which  it  appears  that  equation  (3)  can  be  written  in  the 

f0™        jj^J^t[m  +  if(^)+m].         (4) 

This  is  the  prismoidal  formula. 

If  the  integral  (1)  is  interpreted  as  an  area,  the  result  (4) 
may  be  expressed  as  follows :  The  area  bounded  by  the  axis  of  x, 
two  ordinates,  and  a  curve  may  be  found  approximately  by  multi- 
plying one  sixth  of  the  distance  between  the  ordinates  by  the  sum  of 
the  first,  the  last,  and  four  times  the  middle  ordinate. 

If  the  .integral  (1)  arises  in  finding  the  volume  V  of  a  solid 
with  parallel  bases,  then  formula  (4)  becomes 

F=|(2?  +  4Jf+5),  (5) 

where  h  is  the  altitude  of  the  solid,  B  the  area  of  the  lower 
base,  b  the  area  of  the  upper  base,  and  M  the  area  of  the  section 
midway  between  the  bases. 

Of  course  the  prismoidal  formula  gives  an  exact  result  when 
f(x)  can  be  exactly  represented  in  the  form  (2),  where  any  of 
the  coefficients  may  be  zero.  The  most  important  and  frequent 
cases  in  which  (5)  is  exact  are  those  in  which  f(x)  is  a  quadratic 
polynomial  in  x.  In  this  way  the  student  may  show  that  the 
formula  applies  to  frustra  of  pyramids,  prisms,  wedges,  cones, 
cylinders,  spheres,  or  solids  of  revolution  in  which  the  gener- 
ating curve  is  a  portion  of  a  conic  with  one  axis  parallel  to  the 
axis  of  revolution,  and  also  to  the  complete  solids  just  named. 


APPROXIMATE  INTEGRATION  421 

The  formula  takes  its  name,  however,  from  its  applicability 
to  the  solid  called  the  prismoid,  which  we  define  as  a  solid  hav- 
ing for  its  two  ends  dissimilar  plane  polygons  with  the  same 
number  of  sides  and  the  corresponding  sides  parallel,  and  for 
its  lateral  faces  trapezoids. 

Furthermore,  the  formula  is  applicable  to  a  more  general 
solid  two  of  whose  faces  are  plane  polygons  lying  in  parallel 
planes  and  whose  lateral  faces  are  triangles  with  their  vertices 
in  the  vertices  of  these  polygons. 

Finally,  if  the  number  of  sides  of  the  polygons  of  the  last 
defined  solid  is  allowed  to  increase  without  limit,  the  solid  goes 
over  into  a  solid  whose  bases  are  plane  curves  in  parallel  planes 
and  whose  curved  surface  is  generated  by  a  straight  line  which 
touches  each  of  the  base  curves.  To  such  a  solid  the  formula 
also  applies. 

The  formula  is  extensively  used  by  engineers  hi  computing 
earthworks. 

2.  Simp8orC%  ndr.  When  f(x)  is  not  exactly  expressed  by 
(2),  the  prismoidal  formula  will  in  general  give  better  results 
the  nearer  b  is  to  a.  Hence  we  may  obtain  greater  accuracy 
by  dividing  the  interval  1>  —  <t  into  segments  and  applying  the 
prismoidal  formula  to  each.  Taking  the  interpretation  of  (1) 
as  an  area,  we  divide  the  distance  L  —  <i  into  an  even  number 
(2ri)  of  segments,  each  equal  to  A./-,  and  call  the  values  of  ./• 
at  the  points  of  division  a,  x^  x0,  3*3,  •  •  •,  J\,n_v  b.  At  each  point 
of  division  we  draw  an  ordinate  of  the  curve,  thus  cutting  the 
required  area  into  strips,  and  apply  the  prismoidal  formula  to 
figures  each  of  which  is  made  up  of  two  of  these  strips,  so  that 
SB,  ./•„,  ./'.,  •  •  •,  JC2b_1  correspond  to  the  middle  ordinates  of  these 
figures.     Adding  the  results  thus  obtained,  we  have 


I 


/(sb) dx  =  ^  [/(a)  +  4/0O  +  2/(,; )  +  4/ (,-3)  +  2/(sb4) 


+  --.  +  4./'(  ,-,„_,)  +/(5)].  (6) 

This  is  Simpson's  rule. 

3.  The  trapezoidal  rule.  An  area  may  also  be  computed 
approximately  as  the  sum  of  rectangles,  as  shown  in  §  78.  It  is 
more  exact,  however,  to  replace  the  rectangles  of  fig.  1 25,  §  78,  by 


AC 


422  INFINITE  SERIES 

trapezoids.  This  amounts  to  replacing  a  small  portion  of  the 
curve  y  =f(x)  by  a  straight  line,  which  is  equivalent  to  using 
the  first  two  terms  of  the  series  (2).  If  Ax  and  xv  x2,  a?8,  •  •  • 
are  taken  as  in  §  78,  this  method  leads  to  the  result 

f /(*)  dx  =  ^  [/(a)  +  2/00  +  2M)  +  2/W  +  '  •  • 

+  2/0„_1)+/(J)].  (7) 

This  is  the  trapezoidal  rule.  It  is  evident  that  it  gives  less 
accurate  results  than  those  found  by  Simpson's  rule. 

Ex.    Evaluate    C  (1  +  x2)?  dx. 

1.  By  the  prismoidal  formula. 

/(0)  =  1,    /(|)  =  5.859,    /(3)  =  31.623. 
f3(l  +  x2)?dx  =  g[l  +  4(5.859)  +  31.623]  =  28.030. 

2.  By  Simpson's  rule. 
Take  Ax  =  \.    Then 

/(0)  =  1,        /(' )  =  1.398,        /(l)  =  2.828,        /(|)  =  5.859, 
/(2)  =  11.180,        /(§)  =  19.521,        /(3)  =  31.623. 
f3(l  +  x2)trfx  =  £[1  +  4  (1.398)  +  2  (2.828)  +  4  (5.859) 
+  2  (11.180)  +  4  (19.521)  +  31.623] 

=  27.96. 

3.  By  the  trapezoidal  rule. 

Take  Ax  =  \  and  use  the  previous  calculations. 

f3(l  +  x2)^dx  =  1[1  +  2(1.398)  +  2(2.828)  +  2(5.859) 
+  2  (11.180)  +  2  (19.521)  +  31.623] 
=  28.55. 

196.  The  theorem  of  the  mean.  If  in  the  general  form  of 
Taylor's  series  (1),  §  193,  with  R  in  the  form  (5),  §  193,  we 
take  n  =  1,  we  obtain 

f(x)  =/(«)  +  (x  -«)/'(£>,  (1) 

or,  placing  x  =  a  +•  h, 

/(a  +  A)=/(a)  +  A/'(f),  (2) 

where  £  is  between  a  and  a  +•  h. 


INDETERMINATE  FORMS 


423 


This  result  either  in  the  form  (1)  or  the  form  (2)  is  called  the 
theorem  of  the  mean,  and  has  a  very  simple  graphical  interpre- 
tation. For  let  LK  (fig.  234)  be  the  graph  of  y  =/(»),  and  let 
OA  =  a,OB=a  +  h.    Then  AB  =  h,  f(a)  =  AD,  /(«  +  li)  =  BE, 

and  f(a  +  h)  -f(a) 

h 


the  slope  of  the  chord  BE. 


If  now  £  is  any  value  of  2", /'(£)  is  the  slope  of  the  tangent 
at  the  corresponding  point  of  LK.  Ilenee  (2)  asserts  that  there 
is  some  point  between  I)  and 
E  for  which  the  tangent  is 
parallel  to  the  chord  BE. 
This  is  evidently  true  if  f(x) 
and  f(x)  are  continuous. 

Formula  (1)  may  be  used 
to  prove  the  proposition 
which  we  have  previously 
used  without  proof ;  namely, 
If  the  derivative  of  a  function 
is  always  zero,  the  function  is  a  constant.  For  let /'(./)  be  always 
zero  and  let  a  be  any  value  of  x.  Then,  by  (l),f(x)  —f(a)  =  0. 
That  is,  the  function  is  a  constant. 

From  this  it  follows  that  t/r<>  functions  which  have  tlw  same 
derivative  differ  by  a  constant.  For  if  f'(x)  =  <f>Xx),  then 
d 


</.r 


{f(x)  -  <£(»]  =  0  ;   whence  f(x)  =  tf>(x)  +  C. 

197.  The  indeterminate  form  -.    Consider  the  fraction 
0 


and  let  a  be  a  number  such  that  f(a) 


(1) 
0.    If 


0   and   4>(<t) 
0 

literally  meaningless. 

It  is  customary,  however,  to  define  the  value  of  the  frac- 
tion (1),  when  x=a,  as  the  limit  approached  by  the  fraction 
as  x  approaches  a. 

In  some  cases  this  limit  can  be  found  bv  elementarv  methods. 


424  INFINITE  SERIES 


When  x  —  a  this  becomes  -  •    When  x  5*  a  we  may  divide  both  terms 
of  the  fraction  by  a  —  x,  and  have 

a2  —  x2 
=  a  +  x 


for  all  values  of  x  except  x  =  a.    This  equation  is  true  as  x  approaches  a, 
and  hence 

Lim =  Lim  (a  +  x)  =  2  a. 


Ex.  2. 

x 

When  x  =  0  this  becomes  —    When  i^Owe  have 


1  -  Vl  -  x2      1  -  Vl  -  x2    l  +  Vl-x2 


x  x  i  +  Vl-A-2      1  +  Vl-x2 

„                       T.      1  -  Vl  -  x2      T.                x  n 

Hence  Lim =  Lim =  0. 

x  =  0  X  xi0l  +  Vl-*2 


The  theorem  of  the  mean  may  be  used  to  obtain  a  general 
method.    For  we  have 

f(x)  =  f(a  +  A)  _  f(a)  +  A/'(f ,) 

where  |x  and  f  2  lie  between  a  and  a  +  h.  By  hypothesis, /(a)  =  0, 
4>(a)  =  0.    Therefore  for  h*Q 

As  a:  is  made  to  approach  a,  7i  approaches  zero,  and  |x  and  £2 
approach  a.    Hence 

Lim.zw=m.  (2) 


INDETERMINATE  FOEMS  425 

If,  however,  f'(a)  =  0  and  $'(a)  =  0,  the  right-hand  side  of 
(2)  becomes  -  •  In  this  case  we  take  more  terms  of  Taylor's 
series  and  have  -,n 

whence  Lim  ^  v  J  =  '  „v  y  > 

unless  /"(a)  =  0  and  <£"(«)=  0.    In  the  latter  case  we  take  still 
more  terms  of  Taylor's  series,  with  a  similar  result. 
Accordingly  we  have  the  rule : 

To  find  the  value  of  a  fraction  which  takes  the  form  -  when 
x=  a,  replace  the  numerator  and  the  denominator  each  by  its  deriv- 
ative and  substitute  x  =  a.  If  the  new  fraction  is  also  -  >  repeat 
the  process. 

Ex.  3.    To  find  the  limit  approached  by -when  x  =  0. 

sin  x 

By  the  rule,       Lim  ^  ~  e~*  =  |"fl±iZ!l        =-  =  2. 

x=o      sinx         L    cosx    Jx=o      1 

gSC  —  O  cog  x  "f*  € —  x 

Ex.  4.    To  find  the  limit  approached  by when  x  =  0. 

xsmx 
If  we  apply  the  rule  once,  we  have 

T  .      ex  -  2  cos  x  +  e~x      [ex  +  2  sin  x  —  e~x~\  0 

Lim =    =  -. 

x=o  xsinx  L    sinx  +  xcosx    Jx=o     0 

We  therefore  apply  the  rule  again,  thus : 

T.     e*  —  2cosx  +  e~x      fe*  +  2  cosx  +  e~xl  4      _ 

Lim ; =      — ; =  -  =  2. 

x=o  xsmx  L  2  cosx  —  x  smx  1^=0     2 


198.  Other  indeterminate  forms.  If /(«)  =  oo  and  <£(V)=ao, 
the  fraction  ,  ''  takes  the  meaningless  form  —  when  x  =  a. 
The  value  of  the  fraction  is  then  defined  as  the  limit  approached 


426  INFINITE  SERIES 

by  the  fraction  as  x  approaches  a  as  a  limit.    It  may  be  proved 

that  the.  rule  for  finding  the  value  of  a  fraction  which  becomes  - 

holds  also  for  a  fraction  which  becomes  — . 
.  oo 

The  proof  of  this  statement  involves  mathematical  reasoning 

which  is  too  advanced  for  this  book  and  will  not  be  given. 

Ex.  1.   To  find  the  limit  approached  by     °     (n  >  0)  as  x  becomes  infinite. 

1 

By  the  rule,        Lim     *X  =  Lim =  Lim =  0. 

.t=o)     xn         x=oa  nz"-1      x=oo  nxn 

There  are  other  indeterminate  forms  indicated  by  the  symbols 

0  •  00,  oo  -  00,  0°,  oo°,  1", 

The  form  0  •  oo  arises  when,  in  a  product  f(x)  •  ^>(^)»  we 
have  f(a)=  0  and  <£(V)  =  oo.    The  form  oo  —  oo  arises  when,  in 
f(x)  —  <f>  (V),  we  have  f(a)  =  oo,  <j>  (a)  =  oo. 
-    These  forms  are  handled  by  expressing  f(x)  ■  <f>  (#)  or  f(x)  — 
(f>(x),   as  the   case   may  be,  in   the  form  of  a  fraction   which 

becomes  -  or  ^  when  x  =  a.    The  rule  of  §  197  may  then  be 
applied. 

Ex.  2.   x3e~^. 

xs 
When  x  =  co  this  becomes  co  •  0.   We  have,  however,  x3e_x2  =  — j>  which 

becomes  —  when  x  =  co.    Then 

co 

x3                3  x2                 3  x                   3 
Lim  —  =  Lim =  Lim =  Lim =  0. 

x=«>el!      i=»2i^      x=o,2e^      x=«,±xe3? 

In  the  same  manner  Lim  xne~  ^  =  0  for  any  value  of  n. 


Ex.  3.    sec  x  —  tan  x. 

When  x  =  -  this  is  co  —  co.    We  have,  however, 

1  —  sin  x 

sec  x  —  tan  x  — , 

cos  a; 

which  becomes  -  when  x  =  —  •    Then 
0  2 

Lim  (sec  x  —  tan  x)  =  Lim ■ =  Lim 

.  „.  n         COSX  .  n 


2  2 


FOURIER'S  SERIES  427 

The  forms  0°,  oo°,  1"  may  arise  for  the  function 

[ZOO]**"9  when  »  =  «• 
If  we  place  u  =  [/(a;)  ]*<*>, 

we  have  log  u  =  <f>  (x)  log/(a;). 

If  Lim  <f>  (:r)  log/(.r)  can  be  obtained  by  the  previous  methods, 

x  =  o 

the  limit  approached  by  u  can  be  found. 

i 
Ex.  4.   (1  -  jy. 

When  x  =  0  this  becomes  1".    Place 

then  log  u  =  — 2-i '-  ■ 

x 

now  Limioga-*)  =  r^ri    =_i. 

x  =  0  X  LI  -  xJJ-  =  o 

Hence  log  u  approaches  the  limit  —  1  and  u  approaches  the  limit  -  • 

e 

199.  Fourier's  series.    A  series  of  the  form 

-^  +  « x  cos  ./•  +  a„  cos  2  x  +  •  •  •  +  a„  cos  wa;  -+-  •  •  • 

+  6,  sin  &  +  &a  sin  2x+ h  6„  sin  »./•  H ,  (1) 

where  the  coefficients  aQ,  ax,  •  •  •,  6t,  ?>2,  •  •  •  do  not  involve  a-,  is 
called  a  Fourier's  series.  Every  term  of  (1)  has  the  period  * 
2  7r,  and  hence  (1)  has  that  period.  Accordingly  any  function 
defined  for  all  values  of  x  by  a  Fourier's  series  of  form  (1) 
must  have  the  period  2tt.  But  even  if  a  function  does  not  have 
the  period  2  ir,  it  is  possible  to  find  a  Fourier's  series  which  will 
represent  the  function  for  all  values  of  x  between  —  ir  and  ir, 
provided  that  in  the  interval  —  ir  to  ir  the  function  is  single- 
valued,  finite,  and  continuous  except  for  finite  discontinuities,! 

*f(x)  is  called  a  periodic  function,  with  period  k,  if  f(x  +  k)  =f(x). 

t  If  xx  is  any  value  of  x,  such  that  f(xl  —  e)  and  f(xx  +  e)  have  different 
limits  as  e  approaches  the  limit  zero,  then  f(x)  is  said  to  have  a  finite  discon- 
tinuity for  the  value  x  =  xv  Graphically,  the  curve  y  =  /(x)  approaches  two 
distinct  points  on  the  ordinate  x  =  xv  one  point  being  approached  as  x  increases 
toward  xv  and  the  other  being  approached  as  x  decreases  toward  xv 


428  INFINITE  8EEIES 

and   provided  there  is  not  an   infinite  number  of  maxima  or 
minima  in  the  neighborhood  of  any  point. 

We  will  now  try  to  determine  the  formulas  for  the  coeffi- 
cients of  a  Fourier's  series,  which,  for  all  values  of  x  between 
—  it  and  7r,  shall  represent  a  given  function,  f(x),  which  satisfies 
the  above  conditions. 

Let  f(x)  =  -£  +  a1  cos  x  +  a2  cos  2  x  +  •  •  •  +  an  cos  nx  +  •  •  • 

u 

+  bx  sin  x  +  £„  sin  2  x  +  •  •  •  +  bn  sin  nx  +  •  •  •  •  (2) 

To  determine  aQ,  multiply  (2)  by  dx,  and  integrate  from  —  it 
to  7r,  term  by  term.    The  result  is 

I    f(x)dx  =  a0Tr, 

whence  a  =  —  \    f(x)  dx,  (3) 

since   all  the   terms   on   the   right-hand   side   of  the   equation, 
except  the  one  involving  «0,  vanish. 

To  determine  the  coefficient  of  the  general  cosine  term,  as 
an,  multiply  (2)  by  cos  nxdx,  and  integrate  from  —  -rr  to  tt, 
term  by  term.    Since  for  all  integral  values  of  m  and  n 


L 
£ 


sin  mx  cos  nx  dx  =  0, 


cos  mx  cos  nx  dx  =  0,  (m  =£  n) 

and  /     cos2  nx  dx  =  tt, 

all  the  terms  on  the  right-hand  side  of  the  equation,  except 
the  one  involving  an,  vanish,  and  the  result  is 


£ 


f(x)  cos  nx  dx  =  an7Tj 

T 

i  rn 

whence  an  =  —  I    f(x)  cos  nx  dx.  (4) 

It  is  to  be  noted  that  (4)  reduces  to  (3)  when  n  =  0. 


FOURIER'S  SERIES 


429 


In  like  manner,  to  determine  bn,  multiply  (2)  by  sin  nxdx,  and 
integrate  from  —  ir  to  7r,  term  by  term.    The  result  is 


^JO0 


sin  nx  dx. 


(5) 


For  a  proof  of  the  validity  of  the  above  method  of  deriving  the 
formulas  (3),  (4),  and  (5),  the  reader  is  referred  to  advanced 
treatises. 

Ex.  1.  Expand  x  in  a  Fourier's  series,  the  development  to  hold  for  all 
values  of  x  between  —  tt  and  tt. 


By  (3), 

by  (4), 

and  by  (5), 


1    r* 
i   =  —  I     xdx  =  0, 

TT  J  -  n 
1     /*"■ 

i„  =  -  I      x  cos  nxdx  =  0, 

f>n  =  —  /     x  sin 

7r«/-7r 


nxdx 


2 

-  cos  nir. 


Hence  only  the  sine  terms  appear  in  the  series  for  x,  the  values  of  the 
coefficients  being  determined  by  giving  n  in  the  expression  for  hn  the  values 
1,  2,  :!,•••  in  succession.    Therefore  bl  =  2,  ba  =  —  |,  bs  =  §,  •  •  •,  and 


sin  :?  x 


'(2ir,0)  /(37T,0) 


Tlic  graph  of  the  function  x  is  the  infinite  straight  line  passing  through 
the  origin  and  bisecting  the  angles  of  the  first  and  the  third  quadrant. 

The  limit  curve  of  the  series  coincides  with  this  line  for  all  values  of  x 
between  —  it  and  it,  but  not  for  x  =  —  it  and  x  =  tt;  for  every  term  oi  the 
scries  vanishes  when  x  —  —  tt 
or  x  =  tt,  and  therefore  the 
graph  of  the  series  has  the 
points  (±  tt,  0)  as  isolated 
points  (fig.  235). 

By  taking  xx  as  any  value 
of  x  between  —  tt  and  tt,  and 
giving  k  the  values  1,  2,  3,  •  •  • 

in  succession,  we  can  represent  all  values  of  x  by  xx  ±  2  kir.  But  the  series 
has  the  period  2  tt,  and  accordingly  has  the  same  value  for  xl  ±2  kir  as 
for  a;,.  Hence  the  limit  curve  is  a  series  of  repetitions  of  the  part  between 
x  =  —  tt  and  x  —  tt,  and  the  isolated  points  (±2  k-rr,  0). 

It  should  be  noted  that  the  function  defined  by  the  series  has  finite  discon- 
tinuities, while  the  function  from  which  the  series  is  derived  is  continuous. 


235 


430 


INFINITE  SERIES 


It  is  not  necessary  that/(a:)  should  be  denned  by  the  same  law 
throughout  the  interval  from  —  tt  to  it.  In  this  case  the  integrals 
defining  the  coefficients  break  up  into  two  or  more  integrals,  as 
shown  in  the  following  examples  : 

Ex.  2.  Find  the  Fourier's  series  for  f(x)  for  all  values  of  x  between  —  tt 
and  it,  where  f(x)  =  x  +  tt  if  —  rr  <  x  <  0,  and  /'(./•)  =ir  —  xii.  0<x<ir. 


Here 


a0  =  -  \f    (x  +  tt)  dx  +  j"(v  -  x)  dx\  =  tt  ; 

o„  =  -     /     (./;  +  tt)  cos  nxdx  +  J     (tt  —  x)  cos  nxdx 

2 

=  —  (1  -  cos  rnr) ; 

bn  =  -     J     (x  +  tt)  sin  nx  dx  +  J    (tt  —  x)  sin.  nx  dx\ 


Therefore  the  required  series  is 
4  /cos 


4  /cos x      cos  3  x      cos  ox 
2  +  *  \V~      ~ &~      ~~5*~ 


The  graph  of  f(x)  for  values  of  x  be- 
tween —  it  and  7r  is  the  broken  line  ABC 
(fig.  236).    When  x  =  0  the  series  reduces  to 


&+hh+-)- 


*"b+hh* 


Fig.  230 


When 


±7T 


the  series  reduces  to  0.  Hence  the  limit  curve  of  the  series  coincides  with 
the  broken  line  ABC  at  all  points.  From  the  periodicity  of  the  series  it 
is  seen,  as  in  Ex.  1,  that  the  limit  curve  is  the  broken  line  of  fig.  236. 

Ex.  3.    Find  the  Fourier's  series  for  f(x),  for  all  values  of  x  between  —  tt 
and  tt,  where /(x)  =  0  if  —  tt  <  x  <  0,  &ndf(x)  =  tt  if  0  <  x  <  tt. 

Here 


1    r~ 

an  =  —   f     tt  cos  nx  dx  =  0  ; 

TT  Jo 

1  rn  l 

bn  =  —   I     w  sin  nxdx  =  -  (1  —  cosnrr). 
it  do  nv 

Therefore  the  required  series  is 


7r  ,   n  /sin  a;   ,   sin  3  x  ,   sin  5  x  ,        \ 
2  +  -l-l-  +  — —  +  -*—+--} 

*Byerly,  Fourier'' s  Scries,  p.  40. 


PROBLEMS 


431 


The  graph  of  the  function  for  the  values  of  x  between  —  ir  and  tt  is  the 
axis  of  x  from  x  =—  ir  to  x  =  0,  and  the  straight  line  AB  (fig.  237),  there 
being  a  finite   discontinuity  _r 

when  x  =  0. 

The  curves  (1),  (2),  (3), 
and  (4)  are  the  approxima- 
tion curves  corresponding  re- 
spectively  to   the    equations 


y  =  l>  (i) 

y  =  |  +  2mnx,     (2) 


^4 

/7<o<r^v-x\  „ 

CO 

n 

|\X2) 

^ 

u'¥v 

// 

— 7T     V*/^X 

^-OC 

^JO 

7T        X 

Fig.  237 


sin  5  x\ 
5     / 


(3) 

(1) 


They  may  be  readily  constructed  by  the  method   used   in   §24.     It  is 

to  be  noted  that  all  the  curves  puss  through  the  point   (<»,  -J,  which  is 

midway  between  the  points  A  and  O,  which  correspond  to  the  finite 
discontinuity,  and  that  the  successive  curves  approach  perpendicularity 
to  the  axis  of  x  at  that  point. 


PROBLEMS 


1.  Prove  that  the  series 


1+l+i+i+i+i+l+l+ 


where  there  are  two  terms  of  the  form       i  four  terms  of  the  form 

1       .  1  '""  1 

—  >  eight  terms  of  the  form  —  >   and  2*  terms  of  the  form 


8' 
(k  =  1,  2,  3,  •  •  •),  converges  when  a  >  1. 


(2,)a 


2.  By  comparison   with    the    series    in   problem  1   or  with  the 
harmonic  series   (Ex.  2,   §  187)  prove  that  the  series 


!+*+£+£+ 


+  A  + 

71° 


converges  when  «  >  1,  and  diverges  when  a  ^  1. 


432  INFINITE  SERIES 

By  comparison  with  a  geometric  or  a  harmonic  series  establish 
the  convergence  or  the  divergence  of  the  following  series : 

3.  1  + 1  + 1  + 1  +  .  .  .  +  — L-  +  •  •  •. 

[2       [3       [4  |w  +  1 

2  22  2s  2"_1 

3  3-5       3  ■  5  ■  i  3  ■  5  •  i  •  •  •  (2n  +  1) 

3         4  5  6  to  +  2 

5'  2  +  3T2  +  4T3  +  5T4  +  "*"  +  (n  +  l)n  +  "" 

By  comparison  with  the  series  of  problem  2  establish  the  con- 
vergence of  the  following  series  : 

e.i+|+i+...+_L_+.... 


1  •  3   '   5  •  7   '  9  •  11  '  (4  n  -  3)  (4  n  -  1) 

8  1  I  1  I  1  I 

*  L2-3- 42- 3- 4. 63- 4. 5. 6 

+  rc(»  +  l)(n  +  2)(n  +  3)  +  ""' 

9-i  +  -5  +  i^  +  ---+^TT  +  '-- 

By  the  ratio  test  establish  the  convergence  or  the  divergence  of 
the  following  series  : 

1  1 


10*  2-  l  +  28-  3  +  25-  5H  2a—1(2w-l) 

e       5      5a      53                   5""1 
11.  1 +  -  +  -  +  -  +  ...  + =■+.... 

1       12       [3  |  w.  —  1 

2  22  23  2"     / 

12-i— 2  +  2T-3  +  3^  +  ---+^lf+---' 

13'  5  +  3T52  +  4753+'"  ''(n+l)^"*""' 
12       3  n 

"•S  +  S  +  3  +  -+S  +  -- 


PROBLEMS  433 


-i+f+l+f +•••+£+••••  . 


3       32      33  3" 

\*    1  ,  1    3    ,    1    32  ,  1      3"-1 

16'  2  +  5-2>  +  5>-3>  +  -'-+5^-^r  +  --- 

Find  the  region  of  convergence  of  each  of  the  following  series 

xs        X5  £C2n_1 

17.  a;  +  -  +  -+...  + +  . . .. 

3        5  2  n  —  1 

a;       ar2      *3  ./" 

18-2-2  +  S  +  (P  +  "-  +  (^+--- 

ic  X2  Xs  .r" 

19-  ^r-r>  +  Tr-r  +  r-7i  +  ---  + 


1-2   '  3-  4   '  5-  6   '  '    (2n-l)2n  ' 

1        X        X2  xn~l 

20.  -  +  -2  +  -3+. .•  +  —  +  .... 

03         1     CC8    ,     1     X5    ,  ,    .       ..        ,         1  X2n_1 

21-3-3-3-3  +  5-3^+---+(-1)"-12^i-3^  +  --- 
nn     .      x2  ,  1  •  3    .      1 . 3  •  5  .  , 

22-1--i+TT2xi-rYT3x*  +  --- 

1.3.5-2.-3 
^       '         1.2-3...n-l  +        ' 

Find  the  following  expansions  and  verify  the  given  region  of  ^ 

convergence : 


23.sinx  =  x--+--...+(-ir-  + 

L-  '—  ' (—   GO 


<z<co). 


24.  cosz  =  l-  -  +  -_...  +  (_l).-i__  +  ... 

\=-      li  l±2 (-»<«<  oo). 

25.  log(l  +  aj)=a;-f.  +  f +(-l)-»-  +  •  •  • 

"        (-1<*<1). 

26.  bB^.a(.  +  £  +  £+...  +  ^  +  ...) 

1  —  x        \        3       5  2/1—1  / 

(-l<aj<l). 

27.  tan^  =  ,-f  +  f-...+(-l)-£^+;.. 


434  INFINITE  SERIES 

Expand  each  of  the  following  functions  in  a  series  of  ascending 
powers  of  x,  obtaining  four  terms  no  one  of  which  is  zero : 

28.  .  30.   sec  a;.  32.   log(a:  +  Vl+ar). 
V?14-  x2Y 

31.  e-'seca:.  33.  log  cos  as. 

29.  tana;. 

Find  four  terms  of  the  expansion  into  a  Taylor's  series  of  each 
of  the  following  functions  : 

7T 

34.  cos  a-,  in  the  neighborhood  of  x  =  —  • 

35.  logo-,  in  the  neighborhood  of  x  =  5. 

36.  ex,  in  the  neighborhood  of  as  =  4. 

37.  tan_1a;,  in  the  neighborhood  of  x  =1. 

38.  Vl  -|-  x2,  in  the  neighborhood  of  x  =  2. 

39.  Compute  sin  12°  to  four  decimal  places  by  Maclaurin's  series. 

40.  Compute  sin  46°  to  four  decimal  places  by  Taylor's  series. 

41.  Compute  cos  10°  to  four  decimal  places  by  Maclaurin's  series. 

42.  Compute  cos  32°  to  four  decimal  places  by  Taylor's  series. 

43.  Using  the  result  of  problem  33,  compute  log  cos  18°  to  four 
decimal  places. 

44.  Using  the  series  in  problem  25,  compute  log  §  to  five  decimal 
places. 

45.  Using  the  series  in  problem  26,  compute  log  2  to  five  decimal 
places,  and  thence  by  aid  of  the  result  of  problem  44  find  log  3. 

46.  Using  the  series  in  problem' 2G,  compute  log  |  to  five  decimal 
places,  and  thence  by  aid  of  the  result  of  problem  45  find  log  5. 

47.  Using  the  series  in  problem  26,  compute  log  J  to  four  decimal 
places,  and  thence  by  aid  of  the  result  of  problem  45  find  log  7. 

/        xs     x5  \  M—N 

48.  Prove  logil/=logiV-f  2(  x  +  '—  +  —  +  •  •  •)  where  a;  =  • 

49.  Compute  the  value  of  it  to  four  decimal  places  from  the 

1  7T 

expansion  of  sin_1a;  (Ex.4,  §192)   and  the   relation  sin-1  -  =  —  • 

50.  Compute  the  value  of  it  to  four  decimal  places  from  the  expan- 

1  1         7T 

sion  of  tan-1  a:  (problem  27)  and  the  relation  tan-1  -  +  2  tan  1  -  =  —  • 


PROBLEMS  435 

51.  By  the  binomial  theorem  find  Vl7  to  four  decimal  places. 

52.  By  the  binomial  theorem  find  V26  to  four  decimal  places. 

53.  Show  that   in  the    expansion    of   log  (1  +  x)    (problem   25) 

I'^iSl  when  a>0'  and  '*'<(» +  ixi +  *r'  whe"  x<°' 

1  4-  x 

54.  Show   that,   in   the   expansion    of   log- ■    (problem    26), 

2xn  +  2  X       2\xn  +  2\ 

\R\<7 — T^i ^TTi   when  x>°>    and    \R\<7 — ,    ox/1    i      n»  +  2 

1     '      (n  4-  2)(1  —  x)n  +  2  '     '      (n  4-  2)  (1  +  .*•)"  +  - 

when  x  <  0,  where  n  is  the  exponent  of  x  in  the  last  term  retained 

in  the  expansion. 

55.  By  integrating  the  expansion  of -,  to  obtain  the  expan. 

1+J"  \x"  +  2\ 

sion  of  tan_1a\  show  that  for  the  latter  expansion  \R\  < -;  >  where 

1     '      n  +  2 

n  is  the  exponent  of  x  in  the  last  term  retained  in  the  expansion. 

56.  Show  that,  in  the  expansion  of  (1  +  x)k, 

|  n  +  1 

aml       i^i<Ai^i1(i+4^i^1iwtog<0' 

if  n  -  1:  +  1  >  0. 

57.  From  the  result  of  problem  53  estimate  the  error  made 
in  computing  log  1.2  from  three  terms  of  the  series.  Bow  many 
terms  of  the  series  are  sufficient  to  compute  log  1.2  accurately  to  6 
decimal  places  ? 

58.  From  the  result  of  problem  53  how  many  terms  of  the 
expansion  of  log  (1  +  x)  are  sufficient  to  compute  log  .9  to  5 
decimal  places  ? 

59.  From  the  result  of  problem  54  how  many  terms  of  the  expan- 

1  4-  i- 
sion  of  log are  required  to  compute  log  §  to  4  decimal  places  ? 

60.  Using  the  result  of  problem  55,  find  how  many  terms  of  the 
expansion  of  tan-1a;  are  sufficient  to  compute  tan-1 1  to  four  decimal 
places.  Also  estimate  the  error  made  in  computing  tan-1^  from  5 
terms  of  the  series. 


430  INFINITE  SERIES 

61.  From  the  result  of  problem  56  find  how  many  terms  of  the 
binomial  series  are  sufficient  to  compute  Vl02  to  four  decimal  places. 


62.  Compute  J  ,  approximately, 

(1)  by  the  prismoidal  formula, 

(2)  by  Simpson's  ride,  taking  Ax  =  1, 

(3)  by  the  trapezoidal  rule,  taking  Ax  =  1. 


rz    dx 
7,  (i  +  ^2)2' 


Compute  I    — — ■ — jrj,  approximately, 


(1)  by  the  prismoidal  formula, 

(2)  by  Simpson's  rule,  taking  Ax  =  i, 

(3)  by  the  trapezoidal  rule,  taking  Ax  =  i. 

64.  Compute  J     log  cos  x  dx,  approximately , 

(1)  by  the  prismoidal  formula, 

(2)  by  Simpson's  rule,  taking  Ax  =  —  > 

_ 

(3)  by  the  trapezoidal  rule,  taking  Ax  =  —  • 

Find  the  limit  approached  by  each  of  the  following  functions  as 
the  variable  approaches  its  given  value : 

_  _      2  COS  2  X  —  1  .    7T 

65.    ,X: 


7r  6  sin3x 


66. 


sin  x  —  x        . 

72.   )j  =  0. 

x  —  tan  x 


sin  2  x  cot  5  x 


a2x  _  px 

67.   - >  x  =  0. 

2x 


(*-!)' 


68.^ ^,x^f 

2  sin  x  - 1  6 

e*_e-*_2x 

69.   : >x  =  0. 

x  —  sin  x 

log  sin  - 

70. r-^x  =  7r. 

(x  -  tt)2 


73. 

cotx 

74. 

1-logx2        .  A 
x         >  x  =  U. 

^ 

75. 

log  (X  —  7r) 

_2 i  ,  X  =  7T. 

taaf 

76. 

logx 

>  x  =  oo (n  >  0). 

xn 

77. 

(7r  —  2  x)  tan  x,  x  =  — - 

PROBLEMS  437 

86.  {f  +  £C)«  x  =  oo. 

87.  a'1^,  x=l. 

88.  (cosa-)C8CJr,  x==0. 

89.  (1+  sin  xy,  x  =  0. 

90.  a-J,  x  =  0. 

Expand  each  of  the  following  functions  into  a  Fourier's  series  for 
values  of  x  between  —  w  and  ir  : 

91.   x\  92.   e"*. 

93.  /(aj), where /(x)=—9rif  —  7T<a;<0,and/(aj)='7rifO<a;<7r. 

94.  /(./■),  where/(a;)  =  -  a-  if  -  tt  <  x  <  0,  and/(x)  =  0  if  0  <  x  <  nr. 

95.  f(x),  where /(a-)  ='—ir  if  —  7r  <  ./■  <  0,  and/(.r)  =  x  if  0  <  a-  <  ?r. 

96.  f(x),  where /(a)  =  Oif  —  w <x <  0,  and/(x)  =  ./■-  if  0 <  x  <  it. 


78. 

sin  3  x  esc  5  x,  x  =  0. 

7  9. 

e-arlogfrr,  x  =  oo. 

80. 

a;V_ax,  x  =  go. 

81. 

1               1 

>  X  —  7T. 

a.'  —  7r       tan  a; 

_1 "-.a-ii 

a;  —  1       log  a; 

82. 

83. 

a..™*  xi0i 

84. 

(sina;)ta,,J",  a-  =  0. 

CHAPTER  XVIII 


DIFFERENTIAL  EQUATIONS 

200.  Definitions.  A  differential  equation  is  an  equation  which 
contains  derivatives.  Such  an  equation  can  be  changed  into 
one  which  contains  differentials,  and  hence  its  name,  but  this 
change  is  usually  not  desirable  unless  the  equation  contains  the 
first  derivative  only. 

A  differential  equation  containing  x,  y,  and  derivatives  of  y 
with  respect  to  x,  is  said  to  be  solved  or  integrated  when  a 
relation  between  x  and  y,  but  not  containing  the  derivatives, 
has  been  found  which,  if  substituted  in  the  differential  equa- 
tion, reduces  it  to  an  identity. 

The  manner  in  which  differential  equations  can  occur  in 
practice   and   methods  for  their  integration   are  illustrated  in 

I    ii  n  a 


the  following  examples : 


Ex.  1.    Required  the  curve  the  slope  of  which 
at  auy  point  is  twice  the  abscissa  of  the  point. 

By  hypothesis,      —  =  2  x. 
ax 

Therefore  y  =  x%  +  C.  (1) 


Any  curve  whose  equation  can  be  derived 
from  (1)  by  giving  C  a  definite  value  satisfies 
the  condition  of  the  problem  (fig.  238).  If  it  is 
required  that  the  curve  should  pass  through 
the  point  (2,  3),  we  have,  from  (1), 

3  =  4  +  C;      whence      C=-l, 

and  therefore  the  equation  of  the  curve  is 

»  =  *»-!. 


Fig.  238 


But  if  it  is  required  that  the  curve  should  pass  through  (—3,  10),  we 
have,  from  (1),  10  =  9  +  C;      whence     C  =  l, 

and  the  equation  is  y  =  x2  +  1. 

438 


EXAMPLES 


439 


Ex.  2.  Required  a  curve  such  that  the  length  of  the  tangent  from  any 
point  to  its  intersection  with  OF  is  constant. 

Let  P(x,  y)  (fig.  239)  be  any  point  on  the  required  curve.  Then  the 
equation  of  the  tangent  at  P  is 

Y-y  =  ^(X-x), 
dx 

where  (A',  F)  are  the  variable  coordinates  of  a  moving  point  of  the  tangent, 
(x,  y)  the  constant  coordinates  of  a  fixed  point  on  the  tangent  (the  point 

of  tangency),  and  —  is  derived  from  the, 

as  yet  unknown,  equation  of   the   curve'. 
The  coordinates  of  II,  where  the  tangent 

intersects  OY,  are  then  X 


Y 


dx 


and  the  length  of  PR  is  -\\x-  +  •'-(—)"■ 
he  co 


Representing  by  a  the  constant  length 
of  the  tangent,  we  have 


which  is  the  differentia]   equation 
is  clearly 


(1) 

if   tl>. 


239 


-*/ 


y/cp  -  ./■- 


:Va«  -*-+"  leg 


dx  +  C 


a  ±  V  a-  —  . 


+  C.(2) 


The   arbitrary  constant   C  Bhowa   thai 
there  is  an  infinite  number  of  curves  which 

satisfy  the  conditions  of  the  problem. 
Assuming  a  fixed  value  for  C,  we  see 
from  (1)  and  (2)  that  the  curve  is  sym- 
metrical with  respect  to  OY,  that  x-  cannot 

be  greater  than  a2,  that  —  =  0  and  y  =  C 
dx 

when  x  =  a,  and  that  —  becomes  infinite 
dx 

as  x  approaches  zero. 

From  these  facts  and  the  defining  property  the  curve  is  easily  sketched, 

as  shown  in  fi"\  240.    The  curve  is  called  the  tractrix. 


440 


DIFFERENTIAL  EQUATIONS 


Ex.  3.  A  uniform  cable  is  suspended  from  two  fixed  points.  Required 
the  curve  in  which  it  hangs. 

Let  A  (fig.  241)  be  the  lowest  point,  and  P  any  point  on  the  required 
curve,  and  let  PT  be  the  tangent  at  P.  Since  the  cable  is  in  equilibrium, 
we  may  consider  the  portion  AP  as  a  rigid  body  acted  on  by  three  forces, — 
the.  tension  t  at  P  acting  along  PT,  the  tension  h  at  A  acting  horizontally, 
and  the  weight  of  AP  acting  vertically.  Since  the  cable  is  uniform,  the 
weight  of  AP  is  ps,  where  a-  is  the  length  of  AP  and  p  the  weight  of  the 
cable  per  unit,  of  length.  Equating  the  horizontal  components  of  these 
forces,  we  have  ,         ,       , 


and  equating  the  vertical  components,  we  have 
t  sin  <j>  =  ps. 
From  these  two  equations  we  have 


tan</> 
_  dy 


h  ' 


dx 


O 
Fig.  241 


where  -  =  a,  a  constant. 
P 

This  equation  contains  three  variables,  x,  y,  and  s,  but  by  differentiating 
with  respect  to  x  we  have  (§  91) 


(1) 


cPy  _  ds 

dx2      dx 

■■^m 

the  differential  equation  of  the  required  path. 

To  solve 

(1),  place 

—  =  p.    Then 
dx 

dp 

a—  - 

dx 

(1)  becomes 

=  Vl  +  v\ 

or 

log(j; 

dp       _ 

Vi  +  Pn- 

dx . 
a 

whence 

+  Vl+^2)  = 

:-  +  C. 

(2) 


Since  A  is  the  lowest  point  of  the  curve,  we  know  that  when  x  =  0, 
p  =  0.    Hence,  iu  (2),  C  =  0,  and  we  have 


whence,  since  p 


p  +  Vl  +  p2  =  ea, 

p  =  h\ea-e   «*/; 

e   ")  +  C. 


y  =  lv 


VARIABLES  SEPARABLE  441 

The  value  of  C  depends  upon  the  position  of  OX,  since  y  =  a  +  C  when 
x  =  0.  We  can,  if  we  wish,  so  take  OX  that  OA  =  a.  Then  C"  =  0,  and 
we  have,  finally, 

the  equation  of  the  catenary  (fig.  61,  §  27). 

The  order  of  a  differential  equation  is  equal  to  that  of  the 
derivative  of  the  highest  order  in  it. 

The  simplest  differential  equation  is  that  of  the  first  order 
and  of  the  first  degree  in  the  derivative,  the  general  form  of 
which  is 

or  Mdx+2Tdy  =  Q,  (1) 

where  M  and  N  are  functions  of  %  and  //,  <>r  constants. 

In  the  following  articles  we  shall  consider  some  cases  in 
which  this  equation  can  be  readily  solved. 

201.  The  equation  Mdx  +  Ndy  =  0  when  the  variables  can 
be  separated.     If  the  equation  (1),  §  200,  is  in  the  form 

/^)e7x-+/2Cy)<ty  =  0, 

it  is  said  that  the  variables  are  separated.    The  solution  is  then 
evidently 

Jf1(p)dx+jfi(iy)dy  =  €, 

where  c  is  an  arbitrary  constant. 

The  variables  can  be  separated  if  M  and  N  can  each  be 
factored  mto  two  factors  one  of  which  is  a  function  of  x  alone 
and  the  other  a  function  of  y  alone.  The  equation  may  then 
be  divided  by  the  factor  of  M  which  contains  y  multiplied  by 
the  factor  of  N  wliich  contains  x. 

Ex.  1.    dy=f(x)dx. 

From  this  follows  y=(  f(x)  dx  +  c. 

Any  indefinite  integral  may  be  regarded  as  the  solution  of  a  differential 
equation  with  separated  variables. 


442  DIFFERENTIAL  EQUATIONS 


Ex.  2.    VI  -  f  dx  +  VI  -  x2 dy  =  0. 
This  equation  may  be  written 

dx  dy 


=  0: 


Vl  -  x'2      Vl  -  y2 
whence,  by  integration,  sin-1^  +  sin-1^  =  c.  (1) 

This  solution  can  be  put  into  another  form,  thus :  Let  sin_1x  =  <f>  and 
sin-1?/  =  i//.  Equation  (1)  is  then  <j>  +  \p  =  c,  whence  sin(<£  +  if/)  =  sine; 
that  is,  sin  </>  cos  \p  +  cos  <£  sin  ^  =  k,  where  k  is  a  constant.  But  sin  <f>  =  x, 
sin  \p  =  y,  cos  <f>  =  Vl  -  x2,  cos  ^  =  Vl  —  ?/2 ;  hence  we  have 


xVl-y2  +  yVl-x*  =  k.  (2) 

In  (1)  and  (2)  we  have  not  two  solutions,  but  two  forms  of  the  same 
solution,  of  the  differential  equation.  It  is,  in  fact,  an  important  theorem 
that  the  differential  equation  M dx  +  Ndy  =  0  has  only  one  solution  involv- 
ing an  arbitrary  constant.  The  student  must  be  prepared,  however,  to  meet 
different  forms  of  the  same  solution. 


Ex.  3 

•    (1 

-*2)f  +  ^  = 

ax. 

This 

is  readily  written  as 

(l-x 

*)dy 

+  x  (y  -  a) 

\dx  = 

--0, 

or 

dy          x  i 
<l  —  a      1  - 

Ix 

■X* 

=  0; 

whence, 

by  integration, 

log  (se- 

«)- 

-ilog(l- 

x*)-- 

=  e, 

which  is 

i  the 
may 

same  as 
be  written 

y 

a 

-X2 

—  a  ■ 

=  c, 
=  k 

and  this 

Vl  -  X2. 

202.  The  homogeneous  equation  Mdx  +  Ndy  =  0.  A  polynomial 
in  x  and  y  is  said  to  be  homogeneous  when  the  sum  of  the  expo- 
nents of  those  letters  in  each  term  is  the  same.  Thus  ax2 + bxy  +  cy2 
is  homogeneous  of  the  second  degree,  ax3  +  bx2y  +•  cxy2  +  ey3  is 
homogeneous  of  the  third  degree.  If,  in  such  a  polynomial,  we 
place  y  =  vx,  it  becomes  xnf(y)  where  n  is  the  degree  of  the 
polynomial.    Thus 

ax2-}-  bxy  +  cy2=  x2(a  +  bv  +  cv2~), 
ax3+  bx2y-\-  cxy2  +  ey3=  x3(a  +  bv  +  cv2+  ev3). 


HOMOGENEOUS  EQUATION  443 

This  property  enables  us  to  extend  the  idea  of  homogeneity  to 
functions  which  are  not  polynomials.  Representing  by  f(x,  y) 
a  function  of  x  and  y,  we  shall  say  that  f(x,  y)  is  a  homogeneous 
function  of  x  and  y  of  the  nth  degree,  if,  when  we  place  y  =  vx, 
f(x,  y~)=xnF(y).     Thus  V^  +  y2  is  homogeneous  of  the  first 

degree,  since  vV2 -f  y'2  =  x Vl  +  v'2,  and  log-  is  homogeneous  of 
degree  0,  since  log  -  =  log  v  =  x°  log  v. 

When  M  and  N  are  homogeneous  functions  of  the  same  degree 
the  equation  Mdx  +  NJy=0 

is  said  to  be  homogeneous  and  can  be  solved  as  follows : 

Place  y  =  vx.    Then  dy  =  vdx  +  xdv  and  the  differential  equa- 
tion becomes    ^  ^  dx  +  ggf^  (v  dx  +  sdv)  =  0, 
or  [./;(?')  +  ^200]  J*  +  «»  *>  =  0.  (1) 

If  /j(«)  +  tj/j(v)  =£  0,  this  can  be  written 

*     ./iOO  +  ^O) 

where  the  variables  are  now  separated  and  the  equation  may  be 
solved  as  in  §  201. 

If /jOO  +  w/jOO  =  0,  (1)  becomes  dv  =  0 ;  whence  w  =  6'  and  y  =  ex. 

Ex.   (x-2  —  if)  dx  +  "2  .r//'///  =  0. 
Place  y  =  vx.    There  results 

(1-  v*)dx  +  2v(xdv  +  vdx)  =  0, 
die      2vdv  _ 

Integrating,  we  have  logx  +  log(l  +  r2)  =  c  ; 

whence  x(l  +  r2)  =  c, 

■or  ./•"-  +  f  —  ex. 

203.  The  equation 

(axx  +  bty  +  cj  rfz  +  (a2x  +  b„y  +  c2)  dy  =  0  (I) 

is  not  homogeneous,  but  it  can  usually  be  made  so,  as  follows : 

Place  x  =  x'  +  h,       y  =  y'  +  k.  (2) 

Equation  (1)  becomes 
(ttjx'4-  /;1/+  OjA  +  ^Z;  +  cx)  (/x-'+  («2x  +  \,/+  a2A  +  bjc  +  c2)  r//  =  0.    (3) 


444  DIFFERENTIAL  EQUATIONS 

If,  now,  we  can  determine  h  and  k  so  that 

a1h  +  b1k+c1  =  0^>  (4) 

a2A+62fc+c2  =  0j 

(3)  becomes         (o1.x-/  +  b^f) dx  +  ("2r'  +  h.'/)  "V/  ~  ®> 

which  is  homogeneous  and  can  be  solved  as  in  §  202. 

Now  (4)  cannot  be  solved  if  a-J)2  —  <ij»1  =  0.    In  this  case  —  =  -2  =  k, 
where  k  is  some  constant.    Equation  (1)  is  then  of  the  form      l        1 

(axx  +  \y  +  cx)  rfa;  +  [_k  (axx  +  \y)  +  c2]  J//  =  0,  (5) 

so  that,  if  we  place  a-^x  +  bty  —  x',  (5)  becomes 

(V  +  cjdx  +  (kx'  +  c2)  r//  ~  "1</X  =  0, 

which  is  dx  +  - ,kX+C:2 <lx  =  0, 

("l  —  O-JCjX    +  OjCj  —  UjCg 

and  the  variables  are  separated. 
Hence  (1)  can  always  be  solved. 

204.  The  linear  equation  of  the  first  order.    The  equation 

%+fi(*)y=U^  (i) 

where  fx  (x)  and  f2  (re)  may  reduce  to  constants  but  cannot  con- 
tain y,  is  called  a  linear  equation  of  the  first  order. 

An   equation   of  the   form  M dx  -f  Ndy  =  0  may  be   put  in 

form   (V)   if,   after  transforming   it  to   -—  -\ =  0,  —  can  be 

dx      N  N 

expressed  as  /x  (x)  y  —f2  (x)  ;  that  is,  as  the  difference  of  two 

terms  one  of  which  is  y  multiplied  by  a  function  of  x  and  the 

other  of  which  is  a  function  of  x  only. 

To  solve  (1)  let  .ON 

v  J  y  =  uv,  (2) 

where  u  and  v  are  unknown  functions  of  x  to  be  determined  later 
in  any  way  which  may  be  advantageous.    Then  (1)  becomes 

dv         du       „  „ 

,[|+/iW.]  +  «*_/i(,).  (3) 


LINEAR  EQUATION  OF  FIRST  ORDER  445 

Let  us  now  determine  u  so  that  the  coefficient  of  v  in  (3) 
shall  be  zero.     We  have 

£+/.««-* 

or  —  +fx  (x)  dx  =  0, 

of  which  the  general  solution  is 

log  u  +  I  fx  (x)  dx  =  c. 

Since,  however,  all  we  need  is  a  particular  function  which  will 
make  the  coefficient  of  v  in  (3)  equal  to  zero,  we  may  take  c=Q. 


Then  log  u  =  —  /  f\  (x)  dx, 


-f/i(x)dx  , ,. 

or  u  =  e  J  .  (4) 

With  this  value  of  u,  (3)  becomes 

.-/».-*  _/iWi 

dv        ffi(x)dXj.  ,  N 

whence  v  may  be  found  by  integration.    Substituting  the  values 
of  u  and  v  in  (2),  we  have  the  solution  of  (1). 


Ex.   (1  -  a:2)  -2.  +  xy  =  ax. 
dx 

Dividing  the  equation  by  1  —  x2,  we  have  the  linear  equation 
dy         x  nx 

Substituting  uv  for  y,  we  have 

(du  x        \         dv  ax 

v\ u  )  +  u  —  = (2) 

\  dx      1  -  x2    /         da:      1  -  x2  l  J 


446  DIFFERENTIAL  EQUATIONS 

Placing  the  coefficient  of  v  equal  to  zero,  we  have 

lu  .       x 


dx  +  1  - 

■ — -u  - 
-  x- 

■o, 

or 

du 
"7       1 

xdx    _ 

.  —  a,-'2 

:0; 

whence 

log  u  -  i  log  (1 

-  .r2)  = 

:   0, 

so  that 

u  - 

:  Vl  -  X2. 

Substituting 

this 

value  of  u 

in  (2), 

.7,, 

we  have 

Vl- 

„  (IV 
Xi = 

dx 

ox 
''  1  -  x- ' 

whence 

dv  = 

ax 

-dx 

(3) 

(4) 


(i-x2y 


(5) 

(6) 


and  f  =     ,  +  c 

Vl  -  x1 

Substituting  these  values  of  uv  in  the  equation  y  =  ?/c,  we  have  the 

solution  , 

y  =  a  +  cVl—  x2. 

This  example  is  the  same  as  Ex.  3,  §  201,  showing  that  the  methods  of 
solving  an  equation  are  not  always  mutually  exclusive. 

If  (1)  is  in  the  special  form 

<k-ay=f(x),  (5) 

where  a  is  any  constant,  its  solution  is 

,^  +  ^/(,)4;  (6) 

The  proof  is  left  to  the  student. 

205.   Bernouilli's  equation.    The  equation 

may  be  solved  by  the  same  method  that  was  used  in  solving 
the  linear  equation. 


EXACT  EQUATION  447 


(1) 


Ex.   &-1-. 

dx      x 

=  xy. 

Placing  y  = 

uv,  we  have 

Idu      «\         dv 
\dx       x]         dx 

_    ,.-2,,4,.4_ 

Placing 

dx 

-  -  =  0,  we  find  u 

X 

=  r. 

Substituting 

this  value  of  u  in 

dv 

we  have 

_     ..S   .4  . 

1H  =  ,-^; 
</x 

whence 

3  r3       6 

and,  finally,  since  i 

,  _.v 

1  __,•" 
if         2 

(2) 
(3) 


206.  The   exact  equation  Mdx  +  Ndy  =  0.     If    the    left-hand 

member  of  the  equation 

JTV.r  +  jYtfy  =  0  (1) 

is  an  exact  differential,  <//(./,  y)  (§  170),  that  is,  if 

t  y       dx '  W 

(1)  may  be  written  <//'(./-,  //)  =  0,  (3) 

the  solution  of  which  is  evidently 

/(**)«*  (4) 

In  this  case  (1)  is  called  an  exact  differential  equation. 
The  method  of  solving  (1)  is  evidently  to  lind /(.>•,  y)  as  in 
§  170,  and  set  it  equal  to  a  constant. 

Ex.   (4  a*  +  10  xy*  -  3  >/*)  dx  +  (15  x-;/-  -  12  xy»  +  5  y4)  dy  =  0. 

Here  =  30  xtf  —  12  y3  =  ^—  ,  and   the   equation  is  therefore   exact. 

dy  dx 

Hence  its  solution  is  /'(./•,  //)  =  c,  where 

<Y(*. //)  =  4x3  +  10xy3-3y*  (1) 

8a: 

and  ^('''  v)  =  15  xV  -  12  xy3  +  5  v4.  (2) 


448  DIFFERENTIAL  EQUATIONS 

Integrating  (1)  with  respect  to  x,  we  have 

f(x,  y)  =  x*  +  5  xhf  -  3  xy*  +  F(y). 

Substituting  this  value  in  (2),  we  have 

15  xY  ~  I2  *tf  +  F'  0)  =  15  *V  -  12  xy8  +  5  y* ; 

whence  F' (y)  =  5  y*     and     F(j/)  =  yb. 

Therefore  /(.r,  y)  =  x4  +  5  a%8  —  3  xy*  +  ?/5, 

and  the  solution  of  the  differential  equation  is 

a;4  +  5  xhf  —  3  xif  +  yh  =  c. 

207.  The  integrating  factor.    If  the  equation 

Mdx  +  Ndy  =  0  (1) 

is  not  exact,  i.e.  if  — —  =f  — -  > 

c?/       ox 

it  may  be  proved  that  there  exists  an  infinite  number  of  func- 
tions of  x  and  y  such  that  if  (1)  is  multiplied  by  any  one  of 
them  it  is  made  an  exact  equation.  Such  a  function  is  called 
an  integrating  factor. 

No  general  method  is  known  for  finding  integrating  factors, 
though  the  factors  are  known  for  certain  cases,  and  lists  can 
be  found  hi  treatises  on  differential  equations.  Sometimes  an 
integrating  factor  can  be  found  by  inspection.  In  endeavoring 
to  do  this  the  student  should  keep  in  mind  certain  common 
differentials,  such  as 

d(uv)  =  vdu  +  udv, 
?'  du  —  u  dv 


, ,        ;■  u      vdu  —  udv 
d  tan-1  -  = — , 

V  U    +  V 

, ,      u      v  du  —  u  dv 
d  log  -  = , 

V  uv 

d  (u2  +  v2)  =  2  (u  du  +  v  dvy 


CERTAIN  EQUATIONS  OF  SECOND  ORDER        449 

Ex.   {x2  -  f)  dx  +  2  xydij  =  0. 

We  may  write  this  equation  in  the  form 

x2dx  —  y2dx  +  xdty*)  =  0. 

The  last  two  terms  of  the  left-hand  member  of  the  equation  form  the 
numerator  of  dl  —  J. 

Consequently  we  multiply  the  equation  by  — ,  and  have 

dx +  **<*>- f&^i 

x- 

the  solution  of  which  is  x  +  —  =  c, 

x 

or  x2  +  y2  =  ex. 

It  is  to  be  noted  that  it  is  not  necessary  to  use  the  method  of  §  206  to 
solve  the  equation,  for  when  the  integrating  factor  is  found  by  inspection, 

the  solution  is  at  once  evident. 

208.  Certain  equations  of  the  second  order.  There  are  certain 
equations  of  the  second  order,  occurring  frequently  in  practice, 
which  are  readily  integrated.    These  arc  of  the  four  types: 

We  proceed  to  discuss  these  four  types  in  order: 
1    d*y      tr~\ 

By  direct  integration, 

2; -//»*+* 

y  =  I  lf(x)  dx2  +  cxx  +  c%. 

(Ty 

This  method  is  equally  applicable  to  the  equation  — —  =/(#). 


450  DIFFERENTIAL  EQUATIONS 

Ex.  1.  Differential  equations  of  this  type  appear  in  the  theory  of  the 
bending  of  beams.  Each  of  the  forces  which  act  on  the  beam,  such  as  the 
loads  and  the  reactions  at  the  supports,  has  a  moment  about  any  cross  sec- 
tion of  the  beam  equal  to  the  product  of  the  force  and  the  distance  of 
its  point  of  application  from  the  section.  The  sum  of  these  moments 
for  all  forces  on  one  side  of  a  given  section  is  called  the  bending  moment 
at  the  section.    On  the  other  hand,  it  is  shown  in  the  theory  of  beams 

■pi  r 
that  the  bending  moment  is  equal  to  — — ,  where  E,  the  modulus  of  elas- 
ticity of  the  material  of  the  beam,  and  I,  the  moment  of  inertia  of  the 
cross  section  about  a  horizontal  line  through  its  center,  are  constants,  and 
72  is  the  radius  of  curvature  of  the  curve  into  which  the  beam  is  bent. 

Now,  by  8  10G, 

3  *  dfy 

1  dx* 


R  HS)j 

where  the  axis  of  x  is  horizontal.    But  in  most  cases  arising  in  practice 
-j-  is  very  small,  and  if  we  expand  —  by  the  binomial  theorem,  thus : 


1=W1_3/W         I 
R      dx2[_       2\dx)  J 


C B 


we  may  neglect  all  terms  except  the  first  without  sensible  error.    Hence 

the  bending  moment  is  taken  to  be  El  —  -    This  expression  equated  to 

dx'2 
the  bending  moment  as  defined  above  gives  the  differential  equation  of  the 
shape  of  the  beam.  y 

We  will   apply  this   to  find   the 

shape  of   a  beam  uniformly  loaded      A 

and  supported  at  its  ends.  4 7j 

Let  /  be  the  distance  between  the  ^ 

r  TP      242 

supports,  and  to  the  load  per  foot-run. 

Take  the  origin  of  coordinates  at  the  lowest  point  of  the  beam,  which,  by 
symmetry,  is  at  its  middle  point.  Take  a  plane  section  C  (fig.  242)  at  a 
distance  x  from  0  and  consider  the  forces  at  the  right  of  C.  These  are 
the  load  on  CB  and  the  reaction  of  the  support  at  B.    The  load  on  CB  is 

w  I-  —  x\,  acting  at  the  center  of  gravity  of  CB,  which  is  at  the  distance 

1  (I        \2 

2  ~  X  W\2~  X) 

of  — - —  from  C.    Hence  the  moment  of  the  load  is ~ j  which  is 

taken  negative,  since  the  load  acts  downward.    The  support  B  supports 


CERTAIN  EQUATIONS  OF  SECOND  ORDER        451 

half  the  load,  equal  to  —  •    The  moment  of  this  reaction  about  C  is  therefore 
—  ( -  —  x).    Hence  we  have 


"* 


r~>,    wi (i      \    w(i      \"    w/r~     ,\ 


The  general  solution  of  this  equation  is 

EIy=1%{-8~~ii)+CiX+c*' 

But  in  the  case  of  the  beam,  since,  when  x  =  0,  both  y  and  —  are  0,  we 
have  cx  =  0,  c2  =  0.  dx 


Hence  the  required  equation  is 

to  /Px*      x*\ 
EIy  =  2[-8-  12> 

2  ^y -Ax  dlJ\ 


The  essential  thing  here  is  that  the  equation  contains  —  and 

<Py  dx 

■~,  but  does  not  contain  y  except  implicitly  in  these  derivatives. 

Hence,  if  we  place  -~  =  p,  we  have  -—,  —  -*-  >  and  the  equation 
,  dx  'l.r-      dx 

becomes  -j-=f(x,  />),  which   is   a  differential  equation  of  the 

first  order  in  which  p  and  x  are  the  variables.    If  we  can  find 

p  from  this  equation,  we  can  then  find  y  from  -^-=p.    This 
method  has  been  exemplified  in  Ex.  3,  §  200. 


'•SM*!> 


The  essential  thing  here  is  that  the  equation  contains  -~-  and 

—^i  but  does  not  contain  x.    As  before,  we  place  -£-=p,  but 

now  write  —4  =  ,=,,—?>,'  so  that  the  equation  becomes 
,  dxr      dx     dy  dx        ay 

p  —  =f(y,  jo),  which  is  a  differential  equation  of  the  first  order 
in  which  p  and  y  are  the  variables.  If  we  can  find  p  from  tins 
equation,  we  can  find  y  from  -y-  =  p. 


452  DIFFERENTIAL  EQUATIONS 

Ex.  2.  Find  the  curve  for  which  the  radius  of  curvature  at  any  point 
is  equal  to  the  length  of  the  portion  of  the  normal  between  the  point  and 
the  axis  of  x.  .  r-        /dy\a~]  % 

The  length  of  the  radius  of  curvature  is   ± — (§  106).    The 

equation  of  the  normal  is  (§  87)  — - 

dx  dx2 

This  intersects  OX  at  the  point  [x  +  w  — ,  0 ).   The  length  of  the  normal 

I — 7w  V       dx    ' 

is  therefore  y  -v  1  +  (  — )  • 

The  conditions  of  the  problem  are  satisfied  by  either  of  the  differential 


d2y 
dx* 

21  f 


+  +  &• 


y\l+  2    ■  C1) 


b<m 


E^-nM^- 


y 

dx2 

Placing  —  =  p  and  — \  =  p—-\n  (1),  we  have 
dx  dx'2         dy 

l+P*=Py-; 

dy        p  dp 
whence  —  =  — — ^—z  • 

y       1+p2 

The  solution  of  the  last  equation  is 


y  =  Cj  V 1  +  j)1 ; 

Vw2  -  c  2 
whence  />  =  — 2 L . 

Replacing  p  by  — ,  we  have  l  =  dx. 

p         y  dx  Vfir^ 

Transforming  this  equation  to 

- =  dx 


and  integrating,  we  have 


4 


i-1 


whence  Z/  =  7T\e  ei    +e      Cl  / 


(2) 


CERTAIN"  EQUATIONS  OF  SECOND  ORDER        453 

This  is  the  equation  of  a  catenary  with  its  vertex  at  the  point  (c2,  Cj). 

If  we  place  — -  =  p,  and  — -  —  p  —  in  (2),  we  have 
dx  dx-         ay 

whence  —  =  —± — '-  • 

U       1  +  1>~ 

The  solution  of  this  equation  is  y 


Vl 


whence  p  = 


Replacing  p  by  — »  we  have 
ax 


J^U  =  dx. 


Integrating,  we  have      —  v  c j2  _  y-  =  j:  —  ^  f 
or  (x  —  c„)2  +  //2  =  c2. 

This  is  the  equation  of  a  circle  with  its  center  on  OX. 

If  we  multiply  both  sides  of  this  equation  by  2  '  '  </./•,  we  have 

Integrating,  we  have       ('-^j=j  2/(<y)  tfy  +  cx ; 
whence,  by  separating  the  variables,  we  have 


f 


V2//(y)rfy  +  «i 


#+(?„. 


Ex.  3.  Consider  the  motion  of  a  simple  pendulum  consisting  of  a  parti- 
cle P  (fig.  243)  of  mass  m  suspended  from  a  point  C  by  a  weightless  string 
of  length  /.    Let  the  angle  A  CP  —  6,  where  A  C  is  the  vertical,  and  let 

AC 


454 


DIFFERENTIAL  EQUATIONS 


d2s 
A  J'  =  8.    By  §  93  the  force  acting  in  the  direction  AP  is  equal  to  m — - ; 

dt2 
but  the  only  force  acting  in  this  direction  is  the  component  of  gravity. 
The  weight  of  the  pendulum  being  mg,  its  component  in  the  direction  AP 
is  equal  to  —  mg  sin  0.     Hence  the  differential  equation  of  the  motion  is 
d2s  .     „ 


dt2 


•mg  sin 


We  shall  treat  this  equation  on  the  hypothe- 
sis that  the  angle  through  which  the  pendulum 
swings  is  so  small  that  we  may  place  sin  6  =  6, 

without  sensible  error.  Then,  since  6  =  - ,  the 
equation  becomes 

9 


dh 

dt2 


ds 


Multiplying  by  2  ^  dt  and   integrating,  we 
have 

(ds\*  a 


(I) 


(a'  -  **), 


\dt!        L       I  I 

where  a2  is  a  new  arbitrary  constant.    Separating  the  variables,  we  have 

whence  sin- 1  -  =  -i  /-  (t  —  t0), 

where  t0  is  an  arbitrary  constant.    From  this,  finally, 

s  =  a  sin  -i /-  (t  —  /0). 

The  physical  meaning  of  the  arbitrary  constants  can  be  given.  For  a 
is  the  maximum  value  of  s;  it  is  therefore  the  amplitude  of  the  swing. 
When  t  =  t0,  s  =  0 ;  hence  t0  is  the  time  at  which  the  pendulum  passes 
through  the  vertical. 


209.  The  linear  equation  with  constant  coefficients.  The  differ- 
ential equation 

dn~ly  ,  dy 


dny 

dx11        1  dxn 


!+•• 


+  anV  =/0), 


(1) 


where  «x,  «2,  •  •  •,  an_v  an  are  constants,  and  where  f(x)  is  a 
function  of  x  which  may  reduce  to  a  constant  or  even  be  zero, 
is  called  a  linear  differential  equation  with  constant  coefficients. 


LINEAR  EQUATIONS  455 

To  study  (1)  it  is  convenient  to  express  -j-  by  Dy,  — ~  by 
7;,  oLx  dxr 

LPy,  •  •  •,  ~-  by  D"y,  and  to  rewrite  (1)  in  the  form 

Dny  +  ap"-xy  +  •  •  •  +  %_xDy  +  any  =/(ar), 
or,  more  compactly, 

<JT  +  a^-^  •  •  •  +  o^D  +  a,:)y  =f(x).  (2) 

The  expression  in  parentheses  in  (2)  is  called  an  operator, 
and  we  are  said  to  operate  upon  a  quantity  with  it  when  we 
carry  out  the  indicated  operations  of  differential  ion,  multiplica- 
tion, and  addition.  Thus,  if  we  operate  on  sin  x  with  D3—  2D2 
+  3  D  —  5,  we  have 

(D3—  2D2  -f-  3 D  —  5)  sin  x  =  —  cos  x  +  2  sin  ./•  +  3  c< >s  x  —  5  sin  x 
=  2  cos  a;  —  3  sin  ./•. 

Also,  the  solution  of  (1)  or  (2)  is  expressed  by  the  equation 

*  =  7, +  „i/r-+1..  + „,_,/, +  „/<J>'  <3> 

where  the  expression  on  the  right  hand  of  this  equation  is  not 
to  be  considered  as  a  fraction  but  simply  as  a  symbol  to  ex- 
press the  solution  of  (2).  Thus,  if  (2)  is  the  very  simple 
equation  Dy  =f(x),  then  (3)  becomes 

y=±f(x)=jf(x)dx.  (4) 

In  this  case  —  means  integration  with  respect  to  x.    What  the 

more  complicated  symbol  (3)  may  mean,  we  are  now  to  study. 
210.  The    linear    equation    of    the    first   order    with    constant 
coefficients.     The  linear  equation   of  the  first  order  with  con- 
stant coefficients  is 

or,  symbolically,  (D—a)y=f(x).  (1) 


456  DIFFERENTIAL  EQUATIONS 

The  solution  of  this  equation  is  given  in  §  204.    Hence 

y  =  —L-  f(x)  =  ce°*  +  e-je—fix)  dx.  (2) 

The  solution  (2)  consists  of  two  parts.  The  first  part,  ceax,  con- 
tains an  arbitrary  constant,  does  not  contain  f(x),  and,  if  taken 
alone,  is  not  a  solution  of  (1)  unless  f(x)  is  zero.    The  second 

part,  eax  I  e~axf(x)dx,  contains /(:r),  and,  taken  alone,  is  a  solu- 
tion of  (1),  since  (1)  is  satisfied  by  (2)  when  c  has  any  value, 
including  0.    Hence  e~/W(s)<fc  is  called  particular  inU- 

gral  of  (1),  and,  in  distinction  from  this,  ceax  is  called  the 
complementary  function.  The  sum  of  the  complementary  func- 
tion and  the  particular  integral  is  the  general  solution  (2).  The 
complementary  function  can  be  written  down  from  the  left-hand 
member  of  equation  (1),  but  the  determination  of  the  particular 
integral  requires  integration. 

Ex.  1.    Solve  ^  +  3  y  =  5  xs. 
dx 

This  equation  may  be  written 

(D  +  3)y  =  5x*. 
Hence  the  complementary  function  is  ce~3x.    The  particular  integral  is 

-L^  (5  x*)  =  5  e-**fe**xSdx  =  f  z»  -  gz2  +V  x  -  £f. 

Hence  the  general  solution  is  y  =  ce~3x+  §x3  —  §x2  +  ^-x  —  hj. 

Ex.  2.    Solve  —  +  y  =  sin  x. 
dx 

The  complementary  function  is  cc~x.    The  particular  integral  is 

— - — -  sinx  =  e~x  (  ex  sinxdx  =  ^(sinx  —  cosx). 
D  + 1  J 

Therefore  the  general  solution  is  y  =  ce~x  +  \  (sin  a;  —  cos  a;). 

211.  The  linear  equation  of  the  second  order  with  constant 
coefficients.    The  symbol  (D  —  a~)(D  —  K)y  means  that  y  is  to 


LINEAR  EQUATIONS  457 

be  operated  on  with  D  —  b  and  the  result  operated  on  with 
D  —  a.     Now  (D  —  b~)y  =  -j-  —  by,  and  hence 

<*-><*-»»-£(*-*)-.(*-%■ 

=  (^+JP*>  +  *)*  (1) 

where  p  =  —  (a  +  &),  g  =  aft. 

This  result,  obtained  by  considering  the  real  meaning  of  the 
operators,  is  the  same  as  if  the  operators  D  —  a  and  D  —  b  had 
been  multiplied  together,  regarding  D  as  an  algebraic  quantity. 
Similarly,  we  find 

(i)-6)(2)-a)y  =  [2)3-(a+&)D+a5]y=(i)-a)(D-S)y. 

That  is,  the  order  in  which  the  two  operators  D  —  a  and  D  —  b 
are  used  does  not  affect  the  result. 

Moreover,  if  (lP+pD  +  q)y  is  given,  it  is  possible  to  find  a 
and  b  so  that  (1)  is  satisfied.  In  fact,  we  have  simply  to  factor 
D2  +  pD  +  q,  considering  D  as  an  algebraic  quantity. 

This  gives  a  method  of  solving  the   Linear  equation  of  the 

second  order  with  constant  coefficients.     For  such  an  equation 

has  the  form  r>  , 

ay        'iii  j.,  x 

or,  what  is  the  same  thing, 

(D*+pD  +  q)y=f(x),  (2) 

where  p  and  q  are  constants  and  /(a?)  is  a  function  of  x  which 
may  reduce  to  a  constant  or  be  zero. 
Equation  (2)  may  be  written 

whence,  by  (2),  §  210, 

(D  -  V)y  =  jf—f(?)  =  ^""+  f*.Jr-f(x)to. 


458  DIFFERENTIAL  EQUATIONS 

Again  applying  (2),  §  210,  we  have 

=  c^*+  $*  fe-bx(c'eax+  ear  Ce-a:'f(xyx\dx.   (3) 

There  are  now  two  cases  to  be  distinguished: 
I.  If  a  =£  b,  (3)  becomes 

y^c/x+Cle^+eixCUa-^xCe-^f(^d^dx.    (4) 

II.  If  a  =  5,  (3)  becomes 

y  =  02  +  <v*Q  &ax  +  ?* ffe~  "ZOO  **.  (5) 

In  each  case  the  solution  consists  of  two  parts.  The  one  is 
the  complementary  function  cxeax '  +  c^ehx  or  (<?2  +  cxx)  0°%  involving 
two  arbitrary  constants  but  not  involving  f(x).  It  can  be 
written  down  from  the  left-hand  member  of  the  equation,  and 
is,  in  fact,  the  solution  of  the  equation  (D  —  a)  (D  —  b)  y  —  0. 
The  other  part  of  the  general  solution  is  the  particular  integral, 
and  involves  f(x).  Its  computation  by  (4)  or  (5)  necessitates 
two  integrations. 

Formula  (4)  holds  whether  a  and  b  are  real  or  complex. 
But  when  a  and  b  are  conjugate  complex  it  is  convenient  to 
modify  the  complementary  function  as  follows:  Let  us  place 

a  =  m  +  in,  b  =  m  —  in. 

Then  the  complementary  function  is 

-  e-^e**  +  c,e" £nx)  (§194) 

_  g)na:  J"       ^cog  ^  _|_  £  gm  ^^  _|_  ^  ^cog  nx  __  £  gjn  n£y  j 

__  gjm  £  ^  cog  W;Z.  _j_  ^  gul  ^^  (g) 

where  Cx  =  cx  +  cg,  C2  =  i(cx  —  c2).  Since  cx  and  c2  are  arbitrary 
constants,  so  also  are  Cx  and  C2,  and  we  obtain  all  real  forms  of 
the  complementary  function  by  giving  real  values  to  Cx  and  C2. 


LINEAR  EQUATIONS  459 

The  form  (6)  may  also  be  modified  as  follows:  Whatever 

be  the  values  of  C1  and  C0  we  may  always  find  an  angle  a  such 

C  C 

that  cos  a  =  1      i ,  sin  a  =  '2        .    Then  (6*)  becomes 

Vcf+c*  V^+c;2 

kemjr  cos  (jix  —  a),  (7) 

where  a  and  k  =  v  C\  +  C'22  are  new  arbitrary  constants.   Or  we  may 

—  C  C 

find  an  angle  /3  such  that  sin  ft  =  *   - ,  cos  /3 


Then  (6)  becomes  V^  +  3  V^'i  +  ^ 

£v'"r  sin  (wa:  -  £).  (8) 

Ex.1.   ^  +  5^  +  6y  =  ^. 

This  equation  may  be  written 

(2)  +  2)(D  +  3)y  =  e*. 

The  complementary  function  is  therefore  cxer**  +  c.2e~Ax.    To  find  the 
particular  integral  we  proceed  as  follows: 

1  r  1 

(I)  +  3)  v  =    e  =  e~2x     <*xdx  -  -e* 

v  y,/      D  +  2  J  3 

'-5Ti(i«')--/i*to*-B'1 

Therefore  the  general  solution  is 


Ex.2.    g  +  2^  +  y  =  x. 

This  equation  may  be  written  (J)  +  l)'2//  =  x. 

Therefore  the  complementary  function   is  (cx  +  e2a?)  £-*.   To  fin<l  the 
particular  integral  we  proceed  as  follows: 


(Z)  +  1)y  =  77TT^e"'/' 


/./•  =  X  - 1. 


y  =  ^-^  f>  -  1)  =  er* f(x  -  Y)<*dx  =  x-2. 

Therefore  the  general  solution  is 

.'/  =  Oi  +  V)''-  '+■'•-  2. 

Ex.  3.  Consider  the  motion  of  a  particle  of  unit  mass  acted  on  by  an 
attracting  force  directed  toward  a  center  and  proportional  to  the  distance 
of  the  particle  from  the  center,  the  motion  being  resisted  by  a  force  pro- 
portional to  the  velocity  of  the  particle. 


460  DIFFERENTIAL  EQUATIONS 

If  we  take  s  as  the  distance  of  the  particle  from  the  center  of  force,  the 

ds 

attracting  force  is  —  ks  and  the  resisting  force  is  —  h  —  >  where  k  and  h  are 

positive  constants.    Hence  the  equation  of  motion  is 

or  (D*  +  hD  +  fc)s  =  0.  (1) 

The  factors  of  the  operator  in  (1)  are 

We  have  therefore  to  consider  three  cases  : 
I.  h2  -  4  k<0.   The  solution  of  (1)  is  then 


s  =  e    -  (  C  ,  cos 1  +  C9  sin t  } , 


-i  .    (V4k-h*t 

or  s  =  ae    l  sin  I t  —  pj. 

The  graph  of  s  has  the  general  shape  of  that  shown  in  fig.  62,  §  27.  The 
particle  makes  an  infinite  number  of  oscillations  with  decreasing  ampli- 
tudes, which  approach  zero  as  a  limit  as  t  becomes  infinite. 

II.    A2  -  4  k>  0.    The  solution  of  (1)  is  then 

-(--  ^A2-4iA.  (h   .   V/i3-4t> 


hi      4*. 


The  particle  makes  no  oscillations,  but  approaches  rest  as  t  becomes 
infinite. 

III.  h2  -  4  k  =  0.    The  solution  of  (1)  is 

_h 

s  =  (c1  +  c20e    "• 
The  particle  approaches  rest  as  t  becomes  infinite. 

212.  The   general   linear   equation   with   constant   coefficients. 

The  methods  of  solving  a  linear  equation  of  the  second  order 

with  constant  coefficients  are  readily  extended  to  an  equation 

of  the  ?ith  order  with  constant  coefficients.    Such  an  equation  is 

<fry         <fr~xy  dy  .  ., 

d?  +  a'&^+-"  +  "-^  +  ^=/(^         <*> 

or,  symbolically  written, 

(B"  +  a^"-1  +  •  •  •  +  an_^D  +  an~)y  =f(x).  (2) 


Ex.  1 

dx 


LINEAR  EQUATIONS  463 

dx2      dx        J 

This  may  be  written    (D  +  3)  (D  -  2)  y  =  e4 '. 

The  complementary  function  is  rp<-'"  +  c2e-8*.    To  find  the  particular 
integral,  we  place  .  _    ,  4 .. 

and  substitute  in  the  equation.    We  obtain 
li.k-4'  =e4'. 
To  satisfy  the  equation,  we  must  have 

11.1  =  1,     whence  .1  =  ^\. 
Therefore  the  particular  integral  is 

1   —     1  4   '        I 

and  the  general  solution  is 

y-Ci^x+  c2e-"+  V,'4>- 

Ex.2.    ^  +  ^  =  sin2x. 
</.r'      dx2 

This  may  be  written       l)-(I)  +  1)//  =  sin  2  x. 

The  complementary  function  is  therefore  <■,  +  r.,.r  +  <-.^-x.    To  find  the 
particular  integral,  place 

I  =  Asua  2x  +  B cos 2 x, 
and  substitute  in  the  equation.    We  obtain 

(SB-  1  .1  )sin2x  — (4  B  +  SA  )cos2x  =  sin  2x. 
To  satisfy  the  equation,  we  must  have 

SB-  4  A  =1,     1  B+  s.l  =  0, 
whence  Z>  =  ,',,,     A  =  —  ._,',,. 

Therefore  the  particular  integral  is 

I  =  —  J0  sin  2  .c  +   {\,  cos  2  x, 
and  the  general  solution  is 

H  =  <\  +  V  +  c8e-*  -  -A  sm  2  x  +  to  cos  2  x. 

Ex.3.    $K +  &  =  *«-. 
oar      aa; 

Substituting  y  =  exz,  we  have 

This  may  be  written      (D  +  1)  (£>  +  2)  z  =  x'1.  (2) 


fz  +  :illf  +  2z  =  x>.  (1) 


4D4  DIFFERENTIAL  EQUATIONS 

The  complementary  function  is  therefore  cxe~x  +  c2e~2x.    To  find  the 
particular  integral,  place 

I  =  Ax2  +  Bx  +  C, 

and  substitute  in  (1).    We  obtain 

2  Ax"  +  (6  .1  +  2  S)  .*•  +  (2A  +  BB  +  2  C)  =  a,-2. 
Therefore  2  .4  =  1,  G  .1  +  2  B  =  0,  2  ,1  +  35  +  2  C  =  0, 
whence  A  =  h,     B  =  —  i},     and     C  =  J. 

Hence  the  particular  integral  is 

7"_l~2_3~i7 

and  the  general  solution  of  (1)  is 

z  =  Cle-X  +  c2e~2x  +  ^x2  -  |x  +  \, 
whence  y  =  c\  +  c2e-x  +  e*(^  x2  —  §  a;  +  |). 

This  may  be  written 

(D  - 1) (D  +  2)  (Z>  -  2)y  =  e2*. 

Since  2)  —  2  is  a  factor  of  P  (Z>),  we  place 
I  =  Axe2x, 
and  substitute  in  the  equation.    We  obtain 
4Ae2oc  =  e2x, 
whence  A  =  \  and  I  =  \  xe2x. 

The  general  solution  is 

y  =  Clex  +  c2e-2c  +  c3e2x  +  ^a;e2a;. 

~     -    d?y  , 

Ex.  5.    — -  +  y  —  sin  x. 
dx2 

This  may  be  written  (D2  +  l)y  =  sin  a;. 

By  III,  we  write  I  =  Ax  sin  x  +  i?x  cos  x, 

and  substitute  in  the  equation.    There  results 

—  2B  sin  x  +  2A  cos  a;  =  sin  x. 

1  a: 

Therefore  B  = ,  A  —  0,  and  I  —  —  -  cos  x. 


The  general  solution  is 
-2 


?/  =  Cjc"  +  c2e~ix  —  -  cos  x  =  C1  cos  x  +  C2  sin  a;  —  -  cos  x. 


LINEAR  EQUATIONS  465 

Ex.6,    fir      2|&  +  £  =  **-  +  «-. 

tlx6         ax-      ax 

This  may  be  written    D(D  -  l)2  y  =  xe2x  +  e3x. 

The  complementary  function  is  c1  +  (c2  +  c3x)  ex.  The  particular  integral 
I  is  the  sum  of  I1  and  I2,  where  Ix  corresponds  to  the  term  .<r-',  and  /2 
to  e3 '.    To  find  Iv  place  y  =  e'2xz  in  the  equation 

(D3  -2D2  +  D)y  =  are3 
There  results  (7)3  +  4  Z)2  +  5  Z>  +  2)  z  =  x. 

Placing  2  =  ylx  +  .6,  we  find  /l  =  ^,  ZJ  =  —  §. 
Therefore  A  =  I  (-  *  -  5)  e2*. 

To  find  I2,  we  substitute  y  =  AeZx  in  the  equation 

(D8  — 22^+  /))//  =  e8*. 
We  find  72  =  TV  e3*. 

Hence  I  =\{2x-  5)ea*  +  ,V e3x- 

The  general  solution  of  the  equation  is 

V  =  c,  +  ('•.,  +  c8z)  <■•■  +  J  (2  z  -  5)<  - '  +  ,',,  <  Ba>. 

214.  Systems  of  linear  differential  equations  with  constant 
coefficients.  The  operators  of  the  previous  articles  may  be  em- 
ployed in  solving  a  system  of  two  or  more  linear  differential 
equations  with  constant  coefficients,  when  the  equations  involve 
only  one  independent  variable  and  a  number  of  dependent  vari- 
ables equal  to  the  number  of  the  equations.  The  method  by 
which  this  may  be  done  can  best  be  explained  by  an  example. 

,-.       dx      (Jii 

<lt      dt 

These  equations  may  be  written 

(D-l)x  +  (D-4)y  =  e5t,  (1) 

(D-2)x  +  (D-Z)y  =  e2K  (2) 

We  may  now  eliminate  y  from  the  equations  in  a  manner  analogous  to 
that  used  in  solving  two  algebraic  equations.  We  first  operate  on  (1)  with 
D  —  3,  the  coefficient  of  y  in  (2),  and  have 

(D2  -  4  D  +  3)x  +  (D2  -  IB  +  Yl)y  =  2  e«  (3) 


406  DIFFERENTIAL  EQUATIONS 

since  (D  -  Z)e6i  =  5  e6«  -  3  e6t  =  2  r5'.  We  then  operate  on  (2)  with  D-  4 
the  coefficient  of  y  in  (1),  and  have 

(D-  -(W>+  8)  x  +  (D2  -77)  +  12)  y  =  -  2  e2',  (4) 

since  (D  —  4)e2'  =  —  2  e2'.    By  subtracting  (4)  from  (3)  we  have 

(2 D  -  5)x  =  2  e5«  +  2e2',  (5) 

the  solution  of  which  is     a-  =  cxe**  +  §  e5'  —  2  e2'.  (6) 

Similarly,  by  operating  on  (1)  with  (D  —  2)  and  on  (2)  with  D  —  1,  and 
subtracting  the  result  of  the  first  operation  from  that  of  the  second,  we  have 

(2D-5)y  =  -3e5t  +  e2',        "  (7) 

the  solution  of  which  is       y  =  c2e^'  —  §  eu  —  e2(.  (8) 

The  constants  in  (6)  and  (8)  are,  however,  not  independent,  for  the 
values  of  x  and  y  given  in  (6)  and  (8),  if  substituted  in  (1)  and  (2),  must 
reduce  the  latter  equations  to  identities.  Making  these  substitutions,  we 
have 

l(ci  —  ««)«*'  +  e5t  =  e5t, 

whence  it  is  evident  that  c„  =  cv    Therefore,  replacing  cx  by  c,  we  have 
x  =  ce%1  +  2e5'-2e2', 

y  =  ce^     —  3.  e5«  —  g'2^ 

as  the  solutions  of  the  given  equations. 

215.  Solution  by  series.  The  solution  of  a  differential  equa- 
tion can  usually  be  expanded  into  a  series.  This  is,  in  fact,  an 
important  and  powerful  method  of  investigating  the  function 
defined  by  the  equation.  We  shall  limit  ourselves,  however, 
to  showing  by  examples  how  the  series  may  be  obtained.  The 
method  consists  in  assuming  a  series  of  the  form 

y  =  aoxm  +  ajcm  + 1  +  a2xm  +  '2  +  •  .  • , 

where  m  and  the  coefficients  aQ,  a%,  a2,  •  •  •  are  undetermined. 
This  series  is  then  substituted  in  the  differential  equation,  and 
m  and  the  coefficients  are  so  determined  that  the  equation  is 
identically  satisfied. 


SOLUTION  BY  SERIES  467 

Ex.l.    ,g  +  (,-8)2-2,  =  a 

We  assume  a  series  of  the  form  given  above,  and  write  the  expression 
for  each  term  of  the  differential  equation,  placing  like  powers  of  x  under 
each  other.    We  have  then 

x  -3|  =m(m-l)a0xm-1  +  (m+l)ma1xm+  ■■■+  (m  +  r+l)(m  +  r)ar+1xm+r+  •  ■  • , 


ax 

'"",/"+  ••• 

+  (///  + 0","''",  +  ' 

dx 

—3  mil,,.!-'"- 

L— 8(m+l)0|:*i 

-3(m+r+l)ar+1:C»+' 

-2y  = 

-2a0x'» 

—  2i 

Adding  these  results,  we  have  an  expression  which  must  be  identically 
equal  to  zero,  since  the  assumed  series  satisfies  the  differentia]  equation. 
Equating  to  zero  the  coefficient  of  z*-1,  we  have 

m(m-  l)ao=0.  (1) 

Equating  to  zero  the  coefficient  of  xm,  we  have 

(m +1)0/1-3)0,  +  (m-2)a0  =  0.  (-2) 

Finally,  equating  to  zero  the  coefficient  of  xm  *  r,  we  have  tlie  more  general 

relation  (m  +  r  +  l)(m  +  r-3)ar+]  +  (m  +  r-  2)^  =  0.  (3) 

We  shall  gain  nothing  by  placing  a0  =  0  in  equation  (1),  since  a0xra  is 

assumed  as  the  first  term   of  the  series.    Hence  to  satisfy   (1)  we  must 

have  either  A  , 

m  =  0     or     ///  =  4. 

Taking  the  first  of  these  possibilities,  namely  m  =  0,  we  have,  from  (2), 


and  from   (3),  a_+1  = — a..  (i) 

(r  +  l)(r-3)  l  ; 

This  last  formula  ( 1 )  enables  us  to  compute  any  coefficient,  ar  .  ,.  when  we 
know  the  previous  one,  nr.  Thus  we  find  <i.2  =  —  \&1  =  I  ",,.  o8  =  "•  illu' 
therefore  all  coefficients  after  at  are  equal  to  zero. 

Hence  we  have  as  one  solution  of  the  differential  equation  the  polynomial 

!'i  =  "o(l-  £*  +  I.  ■'■-)■  (:>) 

Returning  now  to  the  second  of  the  two  possibilities  for  the  value  of  m, 
we  take  m  =  4.    Then  (2)  becomes 

5  a,  +  2  a0  =  0, 

/•  +  2 

and  ('■])  becomes  ar  .  ,  = ^,..  ("6") 

(r  +  5)(r  +  l) 


468  DIFFERENTIAL  EQUATIONS 

Computing  from  this  the  coefficients  of  the  first  four  terms  of  the  series, 
we  have  the  solution 

*  =  4'-f*'  +  6V-6^7r'  +  -)-  (7> 

We  have  now  in  (5)  and  (7)  two  independent  solutions  of  the  differential 
('(liiation.    A  more  general  solution  is 

y  =  q.'/i  +  coj2> 

and  this  may  be  shown  to  be  the  most  general  solution. 

(I  V  uV 

Ex.  2.    Lcgendre's  equation.    (1  —  a;2)  — -  —  2  x  —  +  n  (n  +  1)  y  =  0.   , 

il.r-  dx  • 

Assuming  the  general  form  of  the  series,  we  have 

— ~  =  m(m—  l)fl0.r"'~2  +  (m  +  l,)ma12rm— J  +  (m+  2)(m  +  l~)a2xm  +  •••, 

-X*d^=  —m(m-l)a0x»> , 

-2xC-^=  -2ma.jrm-  ■■•, 

dx 

n  (n  +  \)y=  n  (n  +  1)  a0xm  +  •  •  • . 

Equating  to  zero  the  coefficients  of  xm~'2,  xm_1,  and  x'n,  we  have 

m  (m  - 1)  a0  =  0,  (1) 

(m  +  l)ma1  =  0,  (2) 

(m  +  2)  (m  +  l)o,  -  (m  -  n)  (m  +  n  +  1)  a0  =  0.  (3) 

To  find  a  general  law  for  the  coefficients,  we  will  find  the  term  contain- 
ing xm+r~2  in  each  of  the  above  expansions,  this  term  being  chosen  because 
it  contains  ar  in  the  first  expansion.    We  have 

d2>i 

-\-  ■■■  +  (m  +  r)  O  +  r  -  l)arxm+r-2  +  ■■-, 

-  x*j¥  = (m  +  r  -  2)(m  +  r  -  3)a,._2xm  +  r-* , 

-  2  x  <]j-  =  2  (m  +  r-  2)a,._2xm  +  r-* , 

n(n  +l)y  =  •••  +  n(n  +  l)a,._.2xm  +  r-2  +  •••. 

The  coefficient  of  xm  +  r-'2  equated  to  zero  gives 

(m  +  r)  (m  +  r  - 1)  ar  -  (m  -n  +  r-2)  (m  +  n  +  r  —  V)ar_2  =  0.   (4) 


»(»  +1) 

«2=            g        «o5 

w         „x                         (n-r  +  2)(n  +  r- 

and  from  (4),            «r=—  ~ ,   — ~, 

-1) 

By  means  of  (0)  we  determine  the  solution 

SOLUTION  BY  SERIES  469 

We  may  satisfy  (1)  either  by  placing  m  =  0  or  by  placing  m  =  1.    We 
shall  take  m  —  0.    Then  from  (2),  ax  is  arbitrary ;  from  (3), 


(5) 
(6) 


>„(■      (»-l)(»  +  2)  ,3,  (»-l)(»-3)(«  +  2)(n  +  4)  \ 

+  M  |8  [5  _  / 

Since  a0  and  Oj  are  arbitrary,  we  have  in  (7)  the  general  solution 
of  the  differential  equation.  In  fact,  the  student  will  find  that  if 
he  takes  the  value  m  =  l  from  (1),  he  will  obtain  again  the  second 
series  in  (7). 

Particular  interest  attaches  to  the  cases  in  which  one  of  the  series  in 
(7)  reduces  to  a  polynomial.  This  evidently  happens  to  the  first  series 
when  n  is  an  even  integer,  and  to  the  second  series  when  n  is  an  odd 
integer.  By  giving  to  a0  or  a,  such  numerical  values  in  each  case  that  the 
polynomial  is  equal  to  unity  when  x  is  equal  to  unity,  we  obtain  from 
the  series  in  (7)  the  polynomials 

>4 

1\  =  x, 


D       3   ,      1 

p,=l*-\*. 

4      4-2             4-2          4 

■J. 

•  2 

„       0  •  7    ,      „  7  •  5    ,  ,  5 

•3 
.  2 

each  of  which  satisfies  a  Legendre's  differential  equation  in  which  n  has 
the  value  indicated  by  the  suffix  of  P.  These  polynomials  are  called 
Legendre's  coefficients. 

AC 


470  DIFFERENTIAL  EQUATIONS 

Ex.   3.    BesseVs  equation.    xl — -  +  x  —  +  (x~  —  n?)y  =  0. 
ax*         dx 

Assuming  the  series  for  y  in  the  usual  form,  we  have 
x2^  =  m(m  -  l)a0x<»  +  (m  +l)ma1xm+i  +  (m  +  2)  (m  +  l)r,,x"'  +  2  + 

as  ^  =  wfl0xm     +  ( m  +  1 )  axxm  +1  +  (m  +  2)  «2x»  + 2  +  •  •  • , 

—  n2y  =  —  n2a0xm  —  rfa1xm+1  —  n'2a2xm  +  '2  —  •  •  •, 

x2y  —  a0xm  +  2  + 

Equating  to  zero  the  coefficient  of   each  of  the  first  three  powers  of  x, 
we  have 

(w«-n«)a0  =  0,  (1) 

[(B,  +  l)»-„rja1.=  0,  (2) 

[(m  +  2)2-ra2]a2  +  a0  =  0.  (3) 

To  obtain  the  general  law  for  the  coefficients,  we  have 


,d*y 

dx2 

dy 


c2  — |  =  •  •  •  +  (m  +  r)  (m  +  r  -  1)  arxm+r  + 


•••+(?»  +  r)  arxm  + '"  +  •  •  •, 

—  n2y  =  ...  —  ri2arxm  +  r  —  •  •  • , 

z2?/  =  •  •  •  +  n,._2i'm  +  r  +  •  •  •. 

Equating  to  zero  the  coefficient  of  xm  +  r,  we  have 

[(m  +  r)2  -  n2] ar  +  ar_.2  =  0.  (4) 

Equation  (1)  may  be  satisfied  by  m  =  ±  n.   We  will  take  first  m  =  n.   Then 
from  (2),  (3),  and  (4)  we  have 

Oj  =  0,  a 


2(2n  +  2)  r(2n  +  r) 

By  use  of  these  results  we  obtain  the  series 

/  x2  x4  \ 

•A  =  **■  I1  ~  2(2»  +  2)  +  2.4.(2»  +  2)(2»  +  4)  ""'")'  (5) 

Similarly,  by  placing  wi  =  —  n,  we  obtain  the  series 

y2  =  a^"(1  +  2(2n-2)  +  2.4.(2n-2)(2»-4)  +  ---)-        (G) 


PROBLEMS  471 

If,  now,  n  is  any  number  except  an  integer  or  zero,  each  of  the  series 
(5)  and  (6)  converges  and  the  two  series  are  distinct  from  each  other. 
Hence  in  this  case  the  general  solution  of  the  differential  equation  is 

V  =  CiVi  +  '■■j.v- 

If  n  =  0  the  two  series  (5)  and  (6)  are  identical.  If  n  is  a  positive 
integer,  series  (6)  is  meaningless,  since  some  of  the  coefficients  become 
infinite.  If  n  is  a  negative  integer,  series  (">)  is  meaningless,  since  some 
of  the  coefficients  become  infinite.  Hence,  if  n  is  zero  or  an  integer,  we 
have  in  (5)  and  (6)  only  one  particular  solution  of  the  differential  equa- 
tion, and  another  particular  solution  must  1><-  found  before  the  general 
solution  is  known.  The  manner  in  which  this  may  be  done  cannot, 
however,  be  taken  up  here. 

The  series  (5)  and  (6)  with  special  values  assigned  to  a0  define  new- 
transcendental  functions  of  x,  called  Bessel's Junctions.  They  are  important 
in  many  applications  to  mathematical  physics. 

PROBLEMS 
Solve  the  following  equations  : 

1.  x(l-y)dx  +  y(l-x)dy  =  0. 

2.  sec2  yd x  -|-  gob2 xdy  =  0. 

3.  fdx  -  (x2  4-  2  xy)dy  =  0. 

4.  (x  Vj/2  +  f  —  y*)dx  +  xydy  =  0. 

V  V 

5.  [(.r  —  ;/),■'  -j-  x"]dx  +  xc'  ill/  =  0. 

6.  (2  X  +  3  y  -  4)  dx  +  <  3  x  +  y  + 1)  dy  =  0. 

7.  (./•  +  i,  -  .-,),/,•  +  (x  4-  y  -  3)dy  =  0. 

8.  (y  —  x*  —  l)dx  +  xdy=0.     10.  dx  +  (x  —  y)dy  =  0. 

9.  xdy  —  (y  +  x*e?x)dx  =  0.        11.  (a:  +  1  fydx  +(x  +  l  fdy  =  dx. 

12.  (y  +  xy*)dx-dy  =  0. 

13.  (1  +  -rfdy  -  [(1  +  x)y  4-  x'f^lx  =  0. 

14.  (2  x  4-  !,<■■" )  dx  4-  (cos  y  +  xenr) dy  =  0. 

.  1 3  x2  +  ^2  +  2  xf  -  2  £\dx  4  (3  f  +4+2  x*y  4-  2  J  W  =  0. 

1       //    |\  .        I\  y-     \ 


472  DIFFERENTIAL  EQUATIONS 

18.  xdx  +  ydy  =  (x2  +  y2)dx. 

19.  xdy  —  ydx  ==  V.r  —  y*dx. 

20.  ye  vdx  —  (xe  •"  +  y*)di/  =  0. 

21.3  sin  (a-  +  y)  dx  +  [3  sin  (x  -f  y)  —  2y  cos  (x  +  y)~]  dy  =  0. 

22.  x  GOS2ydx  —  esc  xdij  =  0. 

23.  c/ic  +  (x  tan  //  —  sec  y)dy  =  0. 

24.  (,//  -  ■Vx2  +  y2)dx  -  xdy  =  0. 

25.  (a-2  -  ?/8)  tfa;  +  3  xi/dy  =  0. 

26.  (2x  —  5y  +  5) <2as  -f  (4  a;  —  y  + 1) dy  =  0. 


■c-- 

■  (2y 

■?/  WW   *      1Ni^ 

=  0. 

j+ri^+yp+f   yp- 

-  .r2  +  1)  dx  +  (a-2  -  1)  dy  =  0. 

29.  x2  Vl  -  v/Va-  -  if  Vl  -  a;V//  =  0. 

30.  2  a^da;  -  (4  x2  +  y2)  dy  =  0. 

31.  xdx  +  ydy  +  (a-2  +  y*)%(ydx  —  a-rfy)  =  0. 

32.  e*  (a-2  +  //  -f  2  a;)  da:  -f-  2  ye*dy  =  0. 

33.  (3  xy  +  2  e^ajcfa;  -  dy  =  0. 

34.  (2  —  xy)ydx  +  (2  +  xy)xdy  =  0. 

35.  (5  y2  -  6  xy)  dx  +  (6  a:2  -  8  xy  +  y2)  r/y  =  0. 

36.  x(l  +  a'2)f/y  -  [(1  +  a-2)//  +  a7/2]</a-  =  0. 

37.  a- <fa  +  (y  —VxVT~y1)dy  =  0. 

38.  [//  cos  2  ic  +  2  (sin  2  a;)$]  da;  +  sin  2  a;dy  =  0. 

40.  (a'3//3  +  //3)^a-  +  (a-3/  -  a;8)dy  =  0. 

41.  (xy  —  xa)dx  +  dy  =  0. 

42.  (a;  -  y  tan"1  -V/a-  +  x  tan"1  -  dy  =  0. 

43.  (2  a;  -  3  y  +  1)  dx  +  (2x  -  3  y  +  2)  dy  =  0. 

44.  (x  +  2  y)  dx  -  (2  x  -  y)  tfy  =  0. 

45.  (x2  +  y2  +  //)  dx  —  2  xydy  =  0. 

46.  [xy  -  xy  (1  +  a;2)] dx  +  (1  +  a9)  dy  =  0. 


50. 

dx2       x 

51. 

dry              1 

dx2       Vl  -  x2 

5" 

dhj            1 

PROBLEMS  473 

47.  (  xy1 j  e*  Wa:  +  x*ydy  =  0. 

48.  (//  +  x2  -  l)xdx  —  (x2  -  1)  <///  =  0. 

49.  (2  a-//2  -  3  y)dx  +  (2  afy  +  x )  dy  =  0. 


B8. 


a-(IJ+(|)=«. 


59.  — -,  +  -f-  tan  x  =  sin  2.r. 

,lx-        (  1  +  ./•/-  '/.<-       '/.'• 


53.  -— t,  =  sec-".'-. 
car 

60. 

dry      dy 

54.  x— -„  -  -y-  =  x2ec. 
(I.i-       dx 

61. 

&  +  *-■ ln« 

d*y      dy 

55.  x-i  +  -r  +  x  =  o. 

car      ax 

62. 
63. 

^=(1)'^ 

—3-^+^ 

S=*v 

64. 


S+>W^-3-» 


65.  Solve  -yr,  =  -  Vy,  under  the  hypothesis  that,  when  x  =—  1 
,/  =  0  and  ^  =  0. 

66.  Solve  -—^  —  2  if  +  G  //,  under  the  hypothesis  that,  when  x  =  —  j 


7/  =  0  and  -^  =  3. 


67.  Solve 


tfy 


y-^  =  tan3//  -4-  tan  //,  under  the  hypothesis  that,  when 


dy 


x  =  —  >  y  =  0  and  -y1  =  1. 

'/'2>/ 
68.   Solve  -=-j  =  tan  //  +  tan3//,  under  the  hypothesis  that,  when 

n  ""        j   ^      1 

a  =  0,  ii  —  —  and  —  =  1. 


474  DIFFERENTIAL  EQUATIONS 

Solve  the  following  equations  : 

69.  ^--^-0,,  =  6-7x-3*2. 

1 1. 1-        dx 

70.  pL  +  &-at-84*2*     «.£  +  •*  +  «,-*■. 

<l.i-       dx  dx-  dx 

72.  ^  +  4  ,]f  +  4.7/  =  4,-  -  8x  +  2. 
dx2  dx 

73.  a  +  9y  =  6*.  76.  ^-3^=2-6x. 
dec2         *  oar         dx 

74.  ^  -  ^  +  v  =  Xs  +  6.  77.  ^  -  4  «  =  e2*  sin  2x. 

d.,-       dx       •  dx2  J 

75.  a  +  8^+16y  =  2fl--.      78.  ^  +  4^  +  o>  =  20cos3x. 
dar         dx  dx-  dx 

79.  4--4-4^  —  3  ?/  =  2  sin  x  +  cos  2  x. 

dx2  dx 

80.  V?-2^  +  3'/  =  e--rcosx.       82.  t4  +  9y  =  2  sin  3  «• 
dx2         dx        ^  dx2  J 

d2«       „  dy 

84.  — 4  —  7  +1  +  10  //  =  xe2*  +  sin  x. 
dx  dx 

85.  ^{  +  2^  +  4  y=  2x  +  3  e2''. 
dx-  dx 

86.  — ^-2-^  —  Zy  =  xr"  +  cos  2  x. 

dx2  dx  J 

87.  p{  -  4  ^  +  8  y  =  4  x2  - 15  cos  3  x. 
dx-  dx 

88.^  +  2/  =  x3  +  x. 

89.  'p{  +  3'p{-p-3i,  =  3x2  +10  sin  3*. 

dxs  dx-        dx 

90.  ^3-2^-4^  +  8y  =  x<r<+4*2. 
dx8  dx2         dx         ^ 


PEOBLEMS  475 

r/3y       d2//        ,  dy 


d*y      d*y         dy 

■  d?-7t?  +  Adx-^/  =  cos2x- 

cPy       ,  dhi      dy 

.   — -  +  2  — —  2  v  =  5  sin  2  ;•  -I-  1  *>  w2" 


94.  -^-  +  3-^  +  3-^  + y  =  5e*  sin 

dx8  dx-  dx 


■hi 


96-  ^>  +  -7?  +  «  =  -3  +  -- 


r/4y  d*V 

97.  ti  +  534-36v=  20  e3*  cos  3  x, 

d*y         dSi 

98.  -2  +  2  -4  =  2  x  +  25  e    f  sin  2  x. 

r/./'4  dx8 

dSf       ,  d*y 

dx*  dx4 


100. 

dt' 

dy 

dt 

o, 

d*y 

dP 

d  r 

-2  — 
dt 

+  X  =  6'2'. 

101. 

dx 

dt' 

hy  =  e 

', 

x  — 

dy 
Tt  =  e 

-'. 

dx      dy 

-  +  J  +  4y  =  81n3*. 

d*X  ,//, 


d*X        ,/// 

^  +  *        '  *""'    dt3    '    ',//   ^  dt 


102'   7^  +  Si  =  shl  '.  106.   ^  +  2^  +  ^  =  2, 


dx      d*y 

7i  +  M'  =  '-st 

dx       dy 

,*-«.-,+* 

107.   £-„V=0, 

ft-.,-..+l 

g  +  *-a 

476  DIFFERENTIAL  EQUATIONS 

Solve  the  following  equations  by  means  of  series : 

108-  *% +(•'■- **>%-»»  =  a 
io9.(.-^g  +  «4  +  e,-ft 

no.  rfg  +  w*+c— «)ir  =  a 

111.  (1  +  a2)-^  +  £c  ~  -  ny  =  0. 
7  oar         ofcc 

113.  &  +  3^  +  ^  =  0. 

tfar  da: 

114.  Prove  that  any  curve  the  slope  of  which  at  any  point  is 
proportional  to  the  abscissa  of  the  point  is  a  parabola. 

115.  Find  a  curve  passing  through  (0,  —  2)  and  such  that  its 
slope  at  any  point  is  equal  to  three  more  than  the  ordinate  of 
the  point. 

116.  Find  the  curve  the  slope  of  which  at  any  point  is  propor- 
tional to  the  square  of  the  ordinate  of  the  point  and  which  passes 
through  (1,  1). 

117.  Find  the.  curve  in  which  the  slope  of  the  tangent  at  any 
point  is  n  times  the  slope  of  the  straight  line  joining  the  point  to 
the  origin. 

118.  Find  in  polar  coordinates  the  equation  of  a  curve  such 
that  the  tangent  of  the  angle  between  the  radius  vector  and  the 
curve  is  equal  to  minus  the  reciprocal  of  the  radius  vector. 

119.  Find  in  polar  coordinates  the  equation  of  a  curve  such  that 
the  tangent  of  the  angle  between  the  radius  vector  and  the  curve  is 
equal  to  the  square  of  the  radius  vector. 

120.  Find  in  polar  coordinates  the  curve  in  which  the  angle  be- 
tween the  radius  vector  and  the  tangent  is  n  times  the  vectorial  angle. 

121.  A  point  moves  in  a  plane  curve  such  that  the  tangent  to  the 
curve  at  any  point  and  the  straight  line  from  the  same  point  to 
the  origin  of  coordinates  make  complementary  angles  with  the  axis 
of  x.    What  is  the  equation  of  the  curve  ? 


PROBLEMS  477 

122.  Show  that  if  the  normal  to  a  curve  always  passes  through 
a  fixed  point  the  curve  is  a  circle. 

123.  Find  the  curve  in  which  the  perpendicular  from  the  origin 
upon  the  tangent  is  equal  to  the  abscissa  of  the  point  of  contact. 

124.  Find  the  curve  in  which  the  perpendicular  upon  the  tangent 
from  the  foot  of  the  ordinate  of  the  point  of  contact  is  a  constant  a. 

125.  Find  the  curve  in  which  the  length  of  the  portion  of  the 
normal  between  the  curve  and  the  axis  of  x  is  proportional  to  the 
square  of  the  ordinate. 

126.  Derive  the  equation  of  a  curve  such  that  the  sum  of  the  ordi- 
nate at  any  point  on  it  and  the  distance  from  the  point  to  the  axis  of 
x,  measured  along  the  tangent,  is  always  equal  to  a  constanl  </. 

127.  Find  the  polar  equation  of  a  curve  such  that  the  perpendic- 
ular from  the  pole  upon  any  tangent  is  /.-  times  the  radius  vector  of 
the  point  of  contact. 

128.  Find  the  curve  in  which  the  chain  of  a  suspension  bridge 
hangs,  assuming  that  the  load  on  the  chain  is  proportional  to  its 
projection  on  a  horizontal  line. 

129.  Find  the  curve  such  that  the  area  included  between  the 
curve,  the  axis  of  x,  a  fixed  ordinate,  and  a  variable  ordinate  is 
proportional  to  the  difference  between  the  fixed  ordinate  and  the 
variable  ordinate. 

130.  Find  the  curve  in  which  the  area  bounded  by  the  curve,  tin- 
axis  of  x,  a  fixed  ordinate,  and  a  variable  ordinate  is  proporl  kraal  to 
the  length  of  the  arc  which  is  pail  .»t'  the  boundary. 

131.  Find  the  curve  in  which  the  length  of  the  arc  from  a  fixed 
point  to  any  point  P  is  proportional  to  the  square  root  of  the 
abscissa  of  P. 

132.  Find  the  space  traversed  by  a  moving  body  in  the  time  t  if 
its  velocity  is  proportional  to  the  distance  traveled  and  if  the  body 
travels  100  ft.  in  10  sec.  and  200  ft.  in  15  sec. 

133.  In  a  chemical  reaction  the  rate  of  change  of  concentration 
of  a  substance  is  proportional  to  the  concentration  of  the  substance. 
If  the  concentration  is  T%v  when  t  =  0,  and  ^i^  when  t  =  5,  find  the 
law  connecting  the  concentration  and  the  time. 


478  DIFFEKENTIAL  EQUATIONS 

134.  Assuming  that  the  rate  of  change  of  atmospheric  pressure 
p  at  a  distance  h  above  the  surface  of  the  earth  is  proportional  to 
tin'  pressure,  and  that  the  pressure  at  sea  level  is  14.71b.  per  square 
inch  and  at  a  distance  of  1G00  ft.  above  sea  level  is  13.8  lb.  per 
square  inch,  find  the  law  connecting  Ji  and  p. 

135.  The  sum  of  $100  is  put  at  interest  at  the  rate  of  5%  per 
annum,  under  the  condition  that  the  interest  shall  be  compounded 
at  each  instant  of  time.    How  much  will  it  amount  to  in  50  yr.  ? 

136.  If  water  is  running  out  of  an  orifice  near  the  bottom  of  a 
cylindrical  tank,  the  rate  at  which  the  level  of  the  water  is  sinking 
is  proportional  to  the  square  root  of  the  depth  of  water.  If  the  level 
of  the  water  sinks  halfway  to  the  orifice  in  20  min.,  how  long  will 
it  be  before  it  sinks  to  the  orifice  ? 

137.  Find  the  deflection  of  a  beam  fixed  at  one  end  and  weighted 
at  the  other. 

138.  Find  the  deflection  of  a  beam  fixed  at  one  end  and  uniformly 
loaded. 

139.  Find  the  deflection  of  a  beam  loaded  at  its  center  and  sup- 
ported at  its  ends. 

140.  Find  the  curve  whose  radius  of  curvature  is  constant. 

141.  Find  the  curve  in  which  the  radius  of  curvature  at  any  point 
varies  as  the  cube  of  the  length  of  the  normal  between  that  point 
and  the  axis  of  x. 

142.  A  particle  moves  in  a  straight  line  from  a  distance  a  towards 

u 

a  center  of  force  which  attracts  with  a  magnitude  equal  to  ^-    If  the 

particle  was  originally  at  rest,  how  long  will  it  be  before  it  reaches 
the  center  ? 

143.  A  particle  moves  in  a  straight  line  from  a  distance  a  towards 
a  center  of  force  which  attracts  with  a  magnitude  equal  to  fir~i.  If 
the  particle  was  originally  at  rest,  how  long  will  it  be  before  it 
reaches  the  center  ? 

144.  A  particle  begins  to  move  from  a  distance  a  towards  a  fixed 
center  of  force  which  repels  with  a  magnitude  equal  to  /x  times  the 
distance  of  the  particle  from  the  center.  If  its  initial  velocity  is 
V/u."'2,  show  that  the  particle  will  continually  approach,  but  never 
reach,  the  center. 


PROBLEMS  479 

145.  A  particle  moves  along  a  straight  line  towards  a  center  of 
force  which  attracts  directly  as  the  distance  from  the  center.  If  it 
starts  from  a  position  of  rest  a  units  from  the  center,  what  velocity 
will  it  have  acquired  when  it  has  traversed  half  the  distance  to  the 
center  ? 

146.  A  particle  moves  in  a  straight  line  from  a  distance  a  towards 

a  center  of  force  which   attracts  with  a  magnitude  equal  to  — ,, 

r  denoting  the  distance  of  the  particle  from  the  center  of  force.    If 

the  particle  had  an  initial  velocity  of  —=>  how  long  will  it  take  to 

Va 
traverse  half  the  distance  to  the  center  ? 

147.  A  body  moves  through  a  distance  <l  under  the  action  of  a 
constant  force.  Its  initial  velocity  is  p  ,  and  its  final  velocity  is  v .. 
Find  the  time  required. 

148.  A  particle  moves  from  rest  to  a  center  of  force  which  attracts 

with  a  magnitude  equal  to  — ■    Show  that  the  average  velocity  on  the 

first  half  of  its  path  is  to  the  average  velocity  on  the  second  half  in 
the  ratio  7T  —  2  :  7T  +  2. 

149.  Assuming  that  gravity  varies  inversely  as  the  square  of  the 
distance  from  the  center  of  the  earth,  find  the  velocity  acquired  by 
a  body  falling  from  infinity  to  the  surface  of  the  earth. 

150.  Find  the  velocity  acquired  by  ;i  body  sliding  down  a  curve, 
without  friction,  under  the  influence  of  gravity. 

151.  A  bullet  is  fired  horizontally  into  ;i  sand  bank  in  which  the 
retardation  is  equal  to  the  square  root  of  the  velocity.  When  will  it 
come  to  rest  if  the  velocity  on   entering  is   100  it.  per  second'/ 

152.  A  motor  boat  weighing  LOOOlb.  is  moving  in  a  straight  line 
with  a  velocity  of  100  ft.  per  second  when  the  motor  is  shut  off.  If 
the  resistance  of  the  water  is  directly  proportional  to  the  velocity  of 
the  boat,  and  is  equal  to  101b.  when  the  velocity  is  1  ft.  per  second, 
how  far  will  the  boat  move  before  its  velocity  is  reduced  to  25  ft. 
per  second  '.'  How  long  will  it  be  before  this  reduction  of  velocity 
takes  place  ? 

153.  A  particle  is  projected  vertically  upward  from  the  earth's 
surface  in  a  medium  in  which  the  resistance  is  k  times  the  square 
of  the  velocity.     If  vt  is  the  velocity  of  projection  and  va  is  the 


480  DIFFERENTIAL  EQUATIONS 

velocity  with  which  the  particle  returns  to  its  starting  point,  find 
the  value  of  yg  in  terms  of  vt,  k,  and  the  mass  of  the  particle. 

154.  The  force  exerted  by  a  stretched  elastic  string  is  directly 
proportional  to  the  difference  between  its  stretched  length  and  its 
natural  length.  One  end  of  an  elastic  string  of  inconsiderable  mass 
and  of  natural  length  2  ft.  is  fastened  at  a  point  on  the  surface  of  a 
smooth  table.  A  particle  of  mass  ^  lb.  is  attached  to  the  other  end 
of  the  string  and  is  drawn  back  till  the  string  is  stretched  by  an 
amount  1  ft.,  and  is  then  released.  Find  the  time  of  a  complete 
oscillation  of  the  particle  if  a  force  of  \  lb.  is  required  to  stretch 
the  string  to  double  its  natural  length. 

155.  A  particle  of  unit  mass  moving  in  a  straight  line  is  acted 
on  by  an  attracting  force  in  its  line  of  motion  directed  towards  a 
center  and  proportional  to  the  distance  of  the  particle  from  the 
center,  and  also  by  a  periodic  force  equal  to  a  cos  kt.  Determine 
its  motion. 

156.  A  particle  of  unit  mass  moving  in  a  straight  line  is  acted 
on  by  three  forces — an  attracting  force  in  its  line  of  motion  directed 
towards  a  center  and  proportional  to  the  distance  of  the  particle 
from  the  center,  a  resisting  force  proportional  to  the  velocity  of 
the  particle,  and  a  periodic  force  equal  to  a  cos  kt.  Determine  the 
motion  of  the  particle. 

157.  Under  what  conditions  will  the  motion  of  the  particle  in 
problem  156  consist  of  oscillations  the  amplitudes  of  which  become 
very  large  as  the  time  increases  without  limit  ? 


ANSWEKS 

(The  answers  to  some  problems  are  intentionally  omitted.) 

Page  13  CHAPTER  I 

1.  6  +  8V5._  9.(',1).  11    (_2^¥,3|i).  13.(-£,0). 

2.  10  +  2V05.  10.  (,'s.  L'j  ■).  12.  (-3,11). 

Page  14 

14.  (4,  7),  (4,  -  1),  (-  4,  3).  16.    \,  0,  00. 

15.  (lW,  tt)'  (If.  -  ;)•  17.  3  V5,  i  ;  8,  0  ;   V13.  -  }. 

"•  (0,  4).  22.  (£,  41).  25.  (1^,-4),  (5§,  -1). 

19.  (—  r-;,  1  .\).  23.  (3f,  -2}),  (4f,-3f).  26.  (- 4,  -  1).  ' 

20.  (4,  5),  (-  2,  -  3).  24.  (-  1 ,  -  u  J  >.    '  27.  (10,  17). 
Page  15 

28.  (-  3,  2).  29.    §  V5,   ',  \/26,  J,  Vl49.  31.  00,  3.  32.  (16,  -  2). 

Page  18 

59.  02,  -  1,  -  13.  62.    !  \  16,  0,  jV§. 

Page  19 

73.   -  5,  -  -1.  -  1.  0. 

Page  37  CHAPTER  II 

77.  (2,  4).  83.  (-4.-3).  (-3'.  -1§).    88.  (.'„  0),  (2,  3). 

78.  (-  I,  §).  84.  (5,  1!).  89.  (0,1).  (1,  .'.,. 

79.  (3,  4),  (-  is,  -4i).  85.  (2,  },),  (1,  2).  90.  (±  3,  ±4). 

80.  (-3,1).                                  86.  (-1.  -l.\).                          91.  (±2V2,2). 
82.  (-  I  ±  }2VZ,  Y  T  |V5).    87.  (±  4,  ±  2)",  (±  3,  ±1},).     92.  (±  1,  ±  },). 

Page  38 

93.  (0,  0),  (2,  ±  1<).      96.  (±1,  •■;-;).  99.  -  2.07.  102.  2.41. 

94.  (1,  ±  1).  97.  (3,  2£),  (-  1,  1  I).     100.   .40.  2.05.  103.  -  2.52. 

95.  (±2V2,  2).  98.1.40."  101.1.12,3.93. 

Page  45  CHAPTER  III 

1.   (1,7),  (-  5,9),  (2, -4).    2.  x2  +  9*/2  =  5.    3.  x2  +  y*  =  4.    4.  x*-y*=4. 
Page  46 

5.  y3  +  3x2  =  0.  7.  jc-  +  4y  =  0.  9.  4s2  +  9y2  =  l. 

6.  ?y2  =  4  j.  8.  2  x2  -  4  ;/2  =  9.  10.  xy  =  6. 

11.  Sxy  =  7.  15.  (V3,  l),  (l,  -V3),  (l  -  V3,  -  1- V§). 

12.  o&  -  c  16.  (|V2,  iV2),  (2V2,  0),  (1V2,  -|V2). 
14    £l  +  *  _  e  17-  (2,  1),  ('-  2,  -  1),  (-  f,  -  21). 

"  4a      46  18.  x2  +  7y2  =  14. 

417.2  481 


4S: 


ANSWEJJS 


Page 
19. 
20. 
21. 
22. 

Page 
32. 

Page 
14. 
15. 
16. 

Page 
27. 
28. 
29. 
30. 


Page 
43. 
44. 
45. 
49. 
50. 

Page 


64. 

Page 
3. 
4. 
5. 
6. 

Page 
10. 
13. 
14. 
15. 
16. 
17. 


47 

5  x-  +  y2  =  5.  24.  2  xy  ±  40  = 

13  y2-  14  =  0.  25.  3x2  +  2ya 

4 x2  +  9  y"  =  36.  26.  3  x2  -  2  y2 

x*-y*  =  4.  27.  y2  =  2x. 

48 

24x2-15y2  =  200. 

65 

6x  +  a.v  —  a*  =  0.  21. 

Ux  +  4y  -18  =  0.  22. 

2Sx-Uy  +  26=0.  23. 

66 

4x-3y  =  0. 
4x  +  7?/  +  13  =  0. 
4x  +  10  2/  +  1  =  0. 
24  x  -  21y  +  109  =  0. 

40.  (-  2,  4),  tan- 

41.  2x— 9y  +  6 

42.  7  x  +  y  -  25 
67 

8x  +  2/  +  9  =  0,  4x+72/+ll=0. 
y  —  1  =  0,  4x  4-  3y-  11  =  0. 
x-3==0,  4x-3?/-9  =  0. 
§VlO^  jfV34,  4V2. 
ffVl7. 


0. 

:  6. 

=  30. 


10x2  4-  5  ?/2  -  22x  +  4  2/  -  20  =  0. 

205  x2  4-  520?/2  =  4264. 
4  xy  +  13  =  0. 


CHAPTER  V 
1  = 


2x  +  y- 
Gx-4y 
5  x  +  4  •(/ 


25.  12  x  +  3  ?/ -2 

26.  2x  +  6y  —  9  = 


f  24  77 

102/4 

2/  —  5 


-35  = 
1  =  0. 
=  0. 


31.  36 x 

32.  4x- 

33.  2x- 

34.  tan-11. 
fi;  (-3, -5),  tan- 1 
=  0,  7x-  6y  +  21  =  0 
:  0,  x  -  7  y  -  25  =  0. 


(3,  1), 


35. 

tan~ 

112. 

36. 

tan- 

1  3_ 

37. 

tan~ 

15. 

38. 
tan 

tan- 
-J4. 

1    9 

51.  Voi,  ffVeT. 

52.  5x-  2y-  0, 
4x+  5?/—  11 
x  —  1  y  +  11  = 

53.  (2,  -  3). 


54.  T2^Vl3. 

55.  T*7V34. 

56.  2V2. 

57.  9±. 


68 

5x_2?y-4 

(H,  *>■ 

(-  5,  -  3£), 


=  0  ;  2  V29. 
("  21,  If). 


92 

5x  —  Sy  +  4  =  0. 

x+v/+8  =  0,  4x-4  2/-7=0. 
2x+  8;//-  5  =  0,  16x-4j/-l 
21x-  77 y  +  96  =  0,  99 x  +  27  y 

93 

y2  +  \y 


CHAPTER  VI 


0. 

86  =  0. 


65.  (±3  2Vl7,  xT8_VT7). 

66.  (4,  -6),  (2|,  -61). 

67.  (-8,  -11),  (10,  13); 
(- 11,  -  15),  (7,  9). 


7.  2/2-10x  +  25  =  0. 

8.  x2  -  3  y"-  -  4  y  +  4  =  0. 

9.  x2  +  (y  -  3)2  =  ±  x3. 


I). 


2x  +  ll 
91  x2  -  24  xy  +  84  y2  -  364  x  - 

X2  +  y2  _  QX  +  1Q?y  +  18  =  0. 

5  x2  +  5  2/2  +  8  x  -  6  ?/  -  15  =  0 

(-1  +  2V3,  0). 

x2  +  2/2  +  4x-32/  =  0. 


11. 


xz  =  ay. 


12.  x4  +  x22/2 


152  y  +  464  =  0. 


x2  +  y2  +  2  x  -  2  2/  -  6 

x2  +  2/2  ±  2ax  =  0. 

2  x  +  2/  -  3  =  0  ;  §  V5. 


ANSWEES 


483 


27. 


37. 


29.  x2  +  y1  +  3  x  -  4  y  =  0. 

30.  x-  +  y-  —  4  x  +  4  y  +  4  =  0, 

x-  +  j/2  -  20  x  +  20  y  +  100  =  0. 


5  +  240x— 240y  +  225  =  0. 

2  y  +  40  =  0. 


34  //  +  81  =  0. 


Kt 


Page  94 

25.  x2  +  y2-2x-2y  =  0. 

26.  2x2  +  2y2  +  3^-2  =  0. 
5  x2  +  5  y2  ±  9  y  -  80  =  0. 
7x2  +  7  //2  -  19x  +  11  y  -  6  =  0. 
4 x2  +  i  y2  -  60 x  -  GOy  +  225  =  0,  64 x2  +  64 
x2  +  y2-  2x4-  12  j/  +  12  =  0,  x2  +  y2-16x 
x2  +  y2  +  26x  +  16y-32  =  0. 
x2  +  y2  —  6x  -  10  y  +  9  =  0,  x2  +  y2  +  18 x  - 
x2  +  2/2-8x- 12^/4- 48  =  0. 

x2  +  ?/2  +  4x  -  2  //  -  20  =  0,  x2  +  if  +  24  x  -  42  „ 

13  x2  + 13  y2  -  156x  -  52  y  +  295  =  0,  13  x2  4- 13  r  -  52  x  - 104  y  +  259  =  0. 
5,3;   |;  (±4,0). 

95 

iV6,  ^Vf;   I;  (±i.V6,0). 
(0,  ±^V2);  iV2;  (0,  ±  fc). 
(-2,1);   (-5,1),  (1,1); 
iV5;  (-2±V5,  l). 

(*.-*>;  G.-41MM*); 

1x7;   <-.-   ';    |  ^7). 

&2X2  +  a2y2  _  2  „,,,'x  _  0_ 

62x2  +  ahj2  -  2  a2by  =  0. 
i  a/385.  1  VlG5. 


Page 
39. 
40. 
41. 


44. 

45. 

Page 
56. 
57. 
58. 
59. 
60. 
61. 


Page 
70. 
71. 
72. 
73. 
74. 
75. 


46.  x2  +  4  y2  -  4x  +  24y  +  24  =  0. 

47.  36 x1  +  25  if  -  72 x  +  60y  +  60  =  0. 

48.  Sx2  +  9//2  =  180. 

49.  :;./-  +  4//2  =  48. 

50.  <..,•-'  + 8 //2  +  16.y- 04  =  0. 

51.  Bx2  +  25,/2  =  225. 

52.  4x2  +  3,/2  =108. 

53.  9x2  +  25  y2  =  225. 

54.  1 3  <-  +  9 if  - 26 x  —  104  =  0. 

55.  :;<lx2  +  20i/2  =  1125. 


\  N   -■ 

!,  \  29;  (±V29,  0)  ; 
2x  ±_5»/  =  0._ 
£Vl3;  (±Vl3,0); 
2x  + 3w=0. 


96 

3x2+4?y2  =  27.  62. 

9  x2  +  8  if  +  36x  -  48  y  -  20  =  0.  63. 

1  <;.,■-  +  1 5  yi  _  32 x  _  00  y  -  1G4  =  0. 

8x2  +  9y2  =  162.  64. 

J  :  3x2  +  4p2  =  8a2. 

4x2  +  •>,/-  =  4  a2;    V\/5. 

65.  (2,  -  3)  ;  Wl3  ;  (2  ±  Vl3,  -  3)_;  3x  -  2  y  -  12  =  0,  3x  +  2  y  =  0. 

66.  (-  1,  2) ;    \  a  10  ;  (-  1,  2  ±  ^VlO)  ;  y  -  2  =  ±  J  V6  (x  4-  1). 

67.  4  x2  -  20  .v2  -  8  x  -  80  //  -  79  =  0. 

68.  100 x2  -  36y2  +  400 x  +  216  y  +  301  =  0. 

69.  &*x2  -  <fy2  -  2  a^X  =  0. 

97 

7  x2  -  9  y2  =  63.  76.  9  x2  -  16  if  =  20. 

6y*  -  4x2  =  20. 

28  X2  -  36  y2  -  56  x  +  144  y  -  291  =  0. 

9  x2  -  16  y2  —  54  x  +  32  y  -  79  =  0. 

3x2-y2  =  3a2. 

153x2-425y2  =  450. 

81.   9  x2  -  7  y2  =  63  ;  \77  x  ±  16  =  0,  4  x  ±  7  =  0 

83.    ( »„  -  2)  ;  y  +  2  =  0  ;   (2,  -  2)  ;  x  +  1  =  0. 


76.  Ox2 -16?/ 

77.  x2 
79.   ± 


1/3  = 


tan 


(.-OS" 


„V5. 

2-e2 


484  ANSWERS 

Page  98 

84.  (-  1,1);  2.r  +  l=0;  {- },,  {•!);  92.  x2  -  4a-  +  10 y  -  11  =  0. 

16  y  —  19  =  0.  93.  y'2  +  10  y  +  20x  +  65  =  0. 

86.  2?/2  -  12y  +  Ox  =  0.  94.  x2  -  lOx  -  Sy  +  9  =  0. 

87.  12 x2  +  36 x  +  ;/  +  25  =  0.  96.  x2  +  2xy+y*—  16«  +  4y+34=0; 

88.  j/a  =  4px  +  4 p2.  x  +  y  -  3  =  0. 

89.  j/2  =  4px-4pa.  96.  2ft. 

90.  v/2  -  4  y  -  8  x  +  28  =  0.  97.  20  ft. 

91.  x2  +  2x  +  8 2/ -15  =  0. 

Page  99 

98.  27^ ft.  105.  (-1,  2);  (-3,  2),  (1,  2);  (-4,2), 

99.  6  ir  V3  in.  (2,  2) ;  3  x  +  7  =  0,  3  x  -  1  =  0. 

100.  (±VE,  0);_V/5x±9  =  0.  106.  x2  + 2/2-5px  =  0. 

101.  (0,  ±  jVV30);  02/±V30=0.  107.  x2  +  y2  -  2x  -  4  (/  =  0. 

102.  (±  Vl5,  0);  3x±2Vl5  =  0.  108.  x2  +  2/2  -  2x  -  6y  =  0. 

103.  4x2+ 72/2  =  84.  109.  3V7x  -  4?y  +  8  =  0. 

104.  (-If,  -  |);  (-  6|,  -  |),  uo      o  =  (a  +  x)\ 

(H, -§);(- &l  -  $h  '           a~x 

(2^, -i);  12x  +  95  =  0,  m      3=       (2a  +  x)* 

12  x- 55  =  0.  '                   a: 

Page  100 

.,      (x  —  a)2  (2  a  —  x)  4r/3_„T2 

112.  y2  = '-j. -■  116.  y  = 


115.  y  = 


X 

x(a  +  x)2 

2a  +  x 

y\2 
v) 

aV2- 

X 

+  ?/ 

a  V2  + 

X 

—  V 

2  ax2 

4:  a2  +  x2 
118.  Two  straight  lines. 

122.  Circle. 

123.  Circle. 

124.  Circle. 

125.  Concentric  circle. 


x2  +  4  u2 

Page  101 

126.  Straight  line. 

129.  Straight  line.                  135.  Two  parabolas. 

127.  Straight  line. 

132.  Parabola.                         136.  Parabola. 

128.  Circle. 

133.  Parabola.                         137.  Hyperbola. 

138.  Parabola. 

Page  102 

139.  Hyperbola. 

140.  Circle.                141.  Witch.                144.  8pV3. 

Page  114 

CHAPTER  VII 

4  p            4  p 
12.x=^,2,  =  -f 

2  at"-              2  at* 

14.  x  = ,  y  = 

1  +  t2            1+t2 

al) 

u-   1            aht          •           i*    ~       »/r«hi",A    tr        2asin3* 

V&2  +  aH2 

V&2+a2*2                                '"  "  y          cos^ 

ANSWERS  485 

16.  x  =  ±  a  sin  0,  y  —  a  (1  T  sin  0)  tan  0. 

17.  x  =  (a  +  b)  cos  <p  —  h  cos <f>,  y  —  (a  +  b)  sin  0  —  A  sin 0. 

18.  x  =  (b  —  a)  cos  0  +  h  cos 0,  y  =  (b  —  a)  sin  <f>  —  h  sin 0. 

19.  Straight  line. 

20.  x  =  a  (1  —  tan  <f>),  y  =  a  tan  0  (1  —  tan  0)  ;  x2  —  a  (x  —  y). 

Page  115 

21.  x  —  a  cos  0,  ?/  =  a  tan  0  ;  x2  (a2  +  ?/2)  =  a4. 

22.  x  =  a  sec  <p,  y  =  a  sin  0  ;  se2  («-  —  y-)  =  a*. 

23.  x  =  a  esc  0  ctn  0,  y  =  a  esc  0  ;  ?/4  =  «2  (x2  +  */2). 

24.  x  =  a  sin  0  (cos  0  +  sec  0),  y  =  a  cos  0  (cos  0  +  sec  0) ; 
y(x*+  2/2)  =  «(x2  +  2*/2). 

25.  x  =  a(l  —  cos0),  y  —  a  tan  0(1  —  CO8  0). 

26.  x  =  a  sec20,  //  =  a  sec20  tan  0  ;  i8  =  «  (x2  +  y2). 
Page  116 

27.  X  =  (a  —  C  tan  0)  sin20,  y  —  (a  cos  0  —  c  sin  0)  sin  0  ; 
?/(x2  +  y2)  =  x(ay  -  ex). 

28.  X  =  a  sin  2  0,  y  =  2  a  cos  0  ;  //4  =  4  a2  (//2  -  x2). 

29.  x  =  -{a2  +  fc2cos20li  y  = -(a2 tan20  +  i28in0cos0) ; 

«(x-«)(x2  +  ?/2)  =  A-2x2. 

30.  x  =  J,  (a  sin  0  +  6  cos  0),  y  =  £  (<(  ens  <p  +  h  sin  0). 

.,  0  c  /  •  0\ 

31.  x  =  a  cos-   i  i/ =     I  sin  0+  tan     )• 

2    '       2l  2j 

32.  x  =  2  a  cos  2  0,  2/  =  2  a  Ci  is  2  0  tan  0  ;  ,«, 


Page  117 


2  </  +  X 


33.  x  =  a  tan  0.  //  =  a  ens  2  0  ;  y  (a2  +  x2)  =  </  (a2  —  x2). 

35.  2  r2//  COS2a  =  '■'-'./•  Mil  -1  a  —  ;/.r-. 

,      2  v  sin  a  c2   . 

36.  <  =  —      -    ;  x  =  —sin  2  a. 

.'/  !/ 

37.  x=    ()■'-' sin  <t  ens, t  +  iTnsii  v2gh+  o2 sin2 a) ; 

t=     (usina  +  \'2i/h  +  u2sin2a), 

1    .      ,  gb  „    it 

38.  sm1      •  39.      • 

•2  u2  4 

40.  tan-1  —  (u2  +  VV*  +  2  i'V<.  -  g-b~). 
gb 

CHAPTEB  VIII 
Page  127 

34.(«V2,0).  37.^0.0^4-^/1^.    B9.(0,0),r±a 

35 


(l.085o,|Y  V        N4    '' 

(2a,  0),  („v2,  0.  38.  (0,0),  (±  2a,  j). 


,     </2    4/    \      */2     4  ' 


486                                         ANSWERS 

Page  128 

40.  Circle.                                              47.  r2  =  a-  cos  20. 

41.  r  =  a  cos20.                                    48.  r  cos0  = 

a  cos  20. 

42.  r  =  2a  cos3 (9.                                   49.  x  =  4  V3 

cos  2  (9                                       50.  x2  +  y1  -  ay  =  0. 
"cos*? '                                     51.  x2  (x2  +  y*)  =  a  V- 

44.  r2  sin  26*  =  14.                                   52.  (x2  +  y2f 

1  -  4  a2x V2  =  0. 

45.  r  =  8  a  (cos 0  +  sintf).                      53.  (x2  +  j/2  +  ax)2  -  a2(x2  +  y2)  =  0. 

46.  r  =  atan0.                                         54.  (x2  +  yl  - 

-ax)2-62(x2  +  ?/2)  =  0. 

Page  129 

56_  r=  50000000                       n.les 
1  -  cos  9 

62.  Circle. 

63.  Circle. 

57.  1.2  million  miles,  or  4.8  million  miles. 

ce2  cos# 

59.  Parabola. 

1  —  e2  cos2  6 

61.  Straight  line. 

65.  Confocal  conic. 

CHAPTER  IX 
Page  150 

,«         2                                                   6 

,o         1 

10 


11 


12. 


2Va 


20.  Increasing  if  x  >—  3  ;  decreasing  if  x  <  —  3. 

21.  Increasing  if  x  <  0  or  x  >  2  ;  decreasing  if  0  <  x  <  2. 

22.  Increasing  if  x  >—  1 ;  decreasing  if  x  <—  1. 

23.  Increasing  if  —  l<x<0orx>l;  decreasing  if  x  <  —  1  or  0  <  x  <1. 

24.  Increasing  if  x  <  2  or  x  >  3  ;  decreasing  if  2  <  x  <  3. 

25.  Increasing  if  x  <  —  1  or  x  >  3  ;  decreasing  if  —  1  <  x  <  3. 

26.  Upward  if  t  <  8£ ;  downward  if  t  >  3£. 

Page  151 

2l'l<t<4i'.c         r      ,  -,.  33.  (0,0),  (3,  -27). 

28.  Increase  if  x  <  5  ;  decrease  if  x  >  5.  „.    ._  „ '     .    ...    _ ' 

29.  Increase  if  x  <  10  ;  decrease  if  x  >  10.  „  '  ,      J      '   .,'    /n    ,' 

60.    (—  /,   —  rf^),   (U,  i), 

0. 

=  0. 


30.  Increase  if  x  < ; 

V3 

4a 

31.  Increase  if  x  <  —  ; 

3 

decrease  if 
decrease  if 

x> 
x> 

a 

4a 
3  ' 

(2,  -  8|). 

36.  (H,  2TV). 

37.  4x  +  ?/  +  7  = 

38.  3x-y  +  18  = 

32.  s  increases  if  x  >  § 

;  decreases  if  x  <  §. 

41.  54. 

Page  152 

42.  (4§,  55f). 

43.  m,  3). 

44.  tan-i|. 

48. 
49. 
50. 

(1,  _!),(- 1,_^). 
(1,  5),  (-  J,  2/7). 
7z-4y-17  =  0, 

45.  tan-1  f. 

189  x- 108  ?/ -209  =  0. 

46.  8x-2/+  12  =  0, 

51. 

5.8;  5.9;  6. 

216x-27?y-176  = 

:0. 

52. 

.160;   .165. 

47.  x  -  2  ?/  +  9  =  0, 

53. 

.423;  .414. 

27x-54?/-7  =  0. 

I 


ANSWERS 

487 

Page  153 

54.  4.411;  4.566. 

57. 

88|.                   60.  108. 

63. 

2.5. 

55.  6. 

58. 

36.                     61.  6|. 

65. 

i 

56.  |. 

59. 

3Hff.              62.  90,000;  677TV. 

66. 

lOf. 

Page  181 

CHAPTER  X 

1.  6x2  +  18x  +  7. 

5  +  1) 

H            4x(x  +  l) 

2.  0(x  +  2)(2x2  +  8j 

V(4x3  + 6x2-5)2 

3.            2a      . 

2x3  +  x-l 

(x  -  a)2 

Vx4  +  x2-2x 

16x 

"   (x-  +  4)2 

(x3  +  8)2 

2x2(6  -x) 
(4-x)2 

n             6(x  +  2) 

(x2  +  4x  +  l)2 

6          38(x  +  l) 

18.               4X         . 

'  (3x2  +  6x  +  5)2 

V   (Xs  +  l)4 

j2  +  4j.  +  1 

7. -• 

19.  3(6x-8)(8x-  l)(x-l)». 

(X2  +  X  +  I)2 

„    4/0    i        1         1        3 

^\  X3         X*        X* 

9.(8x>-2)(l  +  |). 

Vx  +  1 


20.  :i (_'  iJ  -  l)(4x»  -  10x-  +  2x  +  1). 
2  x8  —  x  +  1 


22. 


10. 


_w4_!\_-j_/i+!y  24.  V 


Vx2  +  1 
x  +  2 

(x2  +  1)§ 
(5x4-7x8+6x2+3x-3)(x2-2x  +  3)* 

(x8  +  \)\ 

1/      1  1 


12.  18x(x  +  l)(2x3  +  Sx8+  (ij- 
lS.  12x2(x3-l)3. 


25.  1- 


8\a  1  +  x 

x 


1-x 


\  x-  -  1 


Page  182 

31. 
32. 
33. 
34. 
35. 

36. 
37. 

38. 

39. 

„.        »       . 

X  +  A    'I1  +  X- 

3x          3 

(1  -  X)  A  X2  -  1 

27. 

(x  +  l)vx2-l 
og      2x2 +  «2         2x 

a2  Va2  +  x2 
2(x  +  2/) 
2x  +  3y2 
2  y4  —  4  x3  y  —  x* 
y*-8xy*  +  x* 
2x  +  y 
x  +  2y 
Vx+~y  +  \7x  -  y 
vx  +  j/  —  Vx  —  y 

x4         4a5x3 
~y*''       "**" 

a2  Va2  +  x2      «"' 
«2 

x2     2xi 

"  (<>'-- x2)f 

«2 
30. 

-2x  +  2/;0. 
x  +  2j/ 

x2Va2  +  x2 

488  ANSWERS 


2ax  2a(4a-3y)  45.  3a:  +  5y  ±  16  =  0 

'  3  y2  -  2  ay '     (3  y  -  2  a)8   '  5x-3?/±4=0. 

41     1-y2    .  2(?/2-l)(4x  +  y)  46.  (±  1,  ±  9). 


2xy-l 

(2X2/- 

42. 

x  —  ly 

+  5  a  =  0  j 

Ix  +  y  ■ 

-  15  ft  =  0. 

43. 

31x  +  £ 

I  y  4.  9  a  =  0 ; 

8x-31 

1/  +  42  ft  =  0. 

44. 

(-2,  - 

8). 

Page 

s  183 

56. 

tan-i3, 

tan-ia. 

57. 

tan-i2. 

58. 

■t    . 
— ,  tan- 

2 

*ft- 

59. 

— ,  tan- 
2 

17. 

Page 

184 

47.  8V2. 

48.  2yxy=  5x£x  —  Sxf. 

49.  xx~  ^x  +  yy~^y  =  o^. 

50.  x^x  +  y{~^y  =  ($• 

51.  x:2x  +  i/fy  —  ftZ/jX  —  axxy  =  ftx^. 


60. 

-,  tan-1 2. 
2 

61. 

tan-1 3. 

62. 

0. 

63. 

0,  tan-1  T2j 

64. 

tan-1  .\ . 

65. 

0,  tan-1  1. 

66. 

tan-1 3. 

67. 

tan-1?-. 

68. 

tan-1V2, 

,5V6 

tan-i 

24 

71.(±?V2,±|V2).     t2.(±-£=,±— £=)•     82.  Vp(pT^). 
\     2  2       /  V      Vft2  +  62         Vft2  +  62/ 

Page  185 

ft2fr2 
85.  a-yxx  —  b^x^y  =  0  ; 


■y/b4x{  +  aty{ 

90.  Upward  if  x  >  J;  downward  if  a;  <  J. 

91.  Upward  if  x  <  —  V2  or  x  >  V2  ;  downward  if  —  V2  <  x  <  V2. 

92.  (-  U,  111).  99,  (2>  o),  (-  I,  913);  (§,  4|f). 

93.  (1,  -  3),  (-  I,  3 i 4 ) .  100.  (-  1,  0),  (3,  -  32);  (1,  -  16). 

94.  (0,  Gftl).  101.  (\,  -  ^);  (1,  0),  (h,  -  TV). 

95.  (±  —  V3,  %  a).  102-  (3.  ~  n);  (°.  16),  (2,  0).' 

V      3  2    ;  /       2  2v/2~\ 

96.  (0,  0).  103.      ±  — ,  T  —=■    ;  (0,  0),  (±  2,  0). 

97.  (0;  0).  V      V3  V3/ 


2        '4       y 


Page  186 

104.  Increase  if  x  >V2ft ;  decrease  if  x  <  V2  ft  (ft  =  given  area). 

105.  Increase  if  x  <  -—  ;  decrease  if  x  >  — - -  (ft.  =  hypotenuse). 

V2  V2 

106.  Increase  if  x  <  -  ;  decrease  if  x  >  -  • 

2  2 

107.  Increase  if  x  <  ^p  ;  decrease  if  x  >  ^p  (p  —  perimeter). 


ANSWERS 


4Si» 


108.  Length  is  twice  breadth. 

109.  12  rd.,  18  rd. 

110.  5  ft. 

Page  187 

114.  4  portions  1  ft.  long  ; 

2  portions  4  ft.  long. 

115.  Breadth  =  —V§, 

3 

depth  =—V6. 

116.  Breadth  =  depth. 

117.  13' r  ft.  long. 

118.  (21,  0). 


111.  Side  of  base  =  20  ft.,  depth  =  10  ft. 

112.  Depth  =  one  half  side  of  base. 

113.  (*,*). 

\6   6/ 


(l^H- 


120.  Height  =  twice  radius  of  base. 

121.  J-. 

V2 

122.  Altitude  =  -V2; 

P      4 
base  =  -  (p  =  perimeter). 
4 


Page  188 

123.  Altitude  =  *  radius  of  sphere. 
124    ^5*.   V^ 

sf  '   sl 

125.  200  cu.  in.;  2547  cu.  in. 
127.  Height   of    rectangle  =  radius 
of  semicircle  ; 
semicircle  of  radius  —  • 

IT 

Page  189 


132.  a 


Ian 


miles  on  laud 


miles  in  water. 


Vn2- 
lm 

Vn2  -  m? 

133.  l^i  hr. 

134.  7hr. 

135.  Area  of  ellipse  =  -   area  of 

rectangle. 


128.  -L  in. 

V3 

129.  "VS. 
3 

130.  8  mi.  from  point  on  bank  near- 

est to  A, 

131.  He  travels  8.',  mi.  on  land. 


136.  Altitude  =  \  \  2  radius  of  semi- 

circle. 

137.  Altitude  =  §   altitude   of   seg- 

ment. 

138.  \  150  mi.  per  hour. 

139.  Velocity  in  still  water  —mi. 

per  hour. 

140.  144  w  cu.  ft.  per  hour. 

141.  (1,3),  (5,  -5). 


Page  190 

142.  2  V?  ft.  per  second. 

143.  4  7r  times  distance  from  vertex. 

144.  .1  in.  per  second. 

145.  .02  in.  per  second. 

146.  .00  cm.  per  second. 

147.  34.9  sq.  in.  per  second. 

148.  Forward  if  t  <  1  or  t  >  5  ; 
backward  if  1  <  t  <  5. 


149.  d  max.  when  t  =  .85  ; 
backward  if  2  <  t  <  4. 

150.  -  v0  (y  =  distance  of  top  of  lad- 

der from  ground,  x  =  dis- 
tance of  bottom  of  ladder 
from  wall). 


490 


ANSWEIIS 


Page  191 

151.  6  COS0  (0  is  the  angle  between 

the  wire  on  which  the  bead 
slides  and  the  straight  line 
drawn  from  the  bead  to  the 
fixed  point). 

152.  281 J  ft.  per  minute. 

153.  150  ft.  per  second. 


154.  y 


8         2V(t2  +  l)4  +  4t2 
x*  +  4  '  "       («2  +  l)2 

155.  When  x  =  |. 

156.  20  ft.  per  second  ; 

10  V5  ft.  per  second  ;  (100,  20). 

157.  .22  ft.  per  second. 

158.  Ellipse;  2  J144~2f- 


Page 

1. 
2. 
3. 

4. 


CHAPTER  XI 
212 

2sin32xcos2x.      11.  xctn(x2  +  a2) esc2 (x2  +  a2) . 

3  sin4  3  x  cos3  3  x.    ,  _      x  .  x 
.  _  ,  12.  ctn6-. 

sm3  ax  cos3  ax.  3 

sin2  2  x. 

cos2(l-2x). 

cost  3  x  sin3  3  x.       14.  sec|  *  Un5  5 . 
sin3  2  x  5  5 

15.  esc  bx  (esc  bx  —  ctn  bx) . 
2 
Vl-4x2 
1 


13.  sec5  -  tan  - . 
2        2 


cos  3  2x 

15 

8. 

sin3(2x  +  1). 

16 

9. 

sec2x(l  +  tanx)2. 

0. 

tan4-. 

17 

19. 


V4x-x2 
2 

(x  +  2)  V2x 

X 

V3-6x2- 
1 

X4 

V  3  x  —  x2 


Vx-x2 


Page  213 


99 

3 

°3 

V8-6x-9x2 

2a 

°4 

x2  +  a2 
1 

01> 

x2  —  4  x  +  5 
1 

^fi 

(x  +  l)Vx2  +  2x 

1 

27. 

Va2  -  x2 
2a2x 

x4  +  a4 

2 
x2  +  I ' 

1 


33. 


35. 


xV9x2-l 

2 

(x  +  2)  Vx2" 

X2- 

+  4x 

1 

(x2  +  1)  Vx4 

+  x2 

1 

+  1 

(x2  +  a;)V4x2  +  4x  — 1 


4a2x 


2Va 


.r*  +  a* 

36.  sin-1  Vl- 

37.  2xtan~1- 


38. 
39. 
40. 
41. 
42. 
43. 
44. 
45. 


2x2-4x  + 

x  +  2 
x2  +  4x  +  3 


4x2-9 

1 

3x2-l 
2-x 

3  — 4x  +  x2 
3 

V9x2  +  2 

X2 

Vx6 


ANSWERS 

491 

46.           X 

50.          6         . 

4  +  5  sin  2  x 

53. 

6x3  +  4x 

x  Va2  +  x2 

Vx4  -  a4 

47.  ctnx. 

51.  2csc2x(ctn2x-l). 

54. 

esc3  2  x. 

48.  3  sec  3  x. 

49.  —  sec  -  • 

2 

52.        *3       • 

(x2  +  4)2 

55. 
56. 

(log  ax)2. 

2     /x2  -  1 
X  \x2  +  l" 

Page  214. 

57.  2  tan-1 2  x. 

65.  iac°82*-  sin2x  •  loga. 

74. 

4 

„     3x  +  2 

66.  (e*  +  er*)*.   s/ 

e-2.r  +  e-2* 

58. 

l  +  2x2 

67.  (ae)b  +  "- 

75. 

0. 

59.  tan- 'ax. 

68.  x2ear. 

76. 

<vetn-1(er  —  1). 
(x  +  l)log(x  +  l) 

60.  sec-1  ax. 

69.  x(a^  +  a-^)2. 

X 

77 

61.  —e~*. 

70.  ^+-2. 

(x-  +  2x)f 

X2 

62.  xe'J+i. 

e«  +  1 

78. 

6x3  +  4x2-24x 
x4—  16 

71.  e"xsinmx. 

63. • 

1  +  x2 

72   a*  log  a 
1+  a2* 

79. 

Va2  -  x2 

64.      * 

81. 

73        a* log  a 
V4ax  —  a2* 

80. 
x). 

Vox 

2  Vl  -  x2  > 

x2  -  a2 

- — - — -  log  (x  +  a  +  Vx2  +  2  a 

(x  +  a)3 

-<*) 

82. 

V2e-      ei*. I-  (eI-e2T)log(2- 

V2  e*  —  e2* 

83. 

.vx'(l  +  logx). 

84. 

2/x4~  +  logx  +  (log.r)-l. 

85. 

^g+logx). 

86.  =  (cos  Vx  •  log  tan  Vx  +  sec  Vx). 

2Vx 


x3  tan  x 

88.  -~— 12  x  tan- »  -  +  a  log  (x2  +  a2)l . 
x2  +  a2  L                a  J 

Page  215 

89  y(l-x2-y2)  92  xy\ogy-y2 

'  x(l  +  x2  +  2/2)'  xylogx-x2 

90-  ~V Qo    y(ysecxy-Iogy) 

y2  —  x  93.  -: -■ 

9L  2sin(x  +  2y)-cos(x  +  2y)  _  X{1~y  SeC Xy) 

2cos(x  +  2y)  —  sin(x  +  2y)'  94.  —  e'J~x;  e2»~x. 


492  ANSWERS 

95<  s  +  y.  2(x2+y2) 
'  x-y'     (x  -  y)s 
%    2z  +  y     10(x2  +  s/2) 

'  x  -  2  y  '     (x  —  2  yf 

tan  x         tan2  y  sec2  x  +  tan2  x  sec2  y 


104.  tan-i3,  tan-i^. 


97. 


tan  y '  tan3  y 


105. 

('• 

i)- 

106. 

(* 

1 

V2' 

1 

107. 

e- 

a- 

108. 

6- 

.), 

4. 

98.         y2        ;  tltlM. 
'  x(y  —  x)      x  (y  -  x)3 

102.  tan-!2V2. 

103.  tan-if. 

Page  216 

110.  Turning  points  when  x  =  ±  -1  V2  ; 

points  of  inflection  when  x  =  0  or  ±  1  V6. 

111.  Turning  points  when  x  =  0  or  2  ;  points  of  inflection  when  x  =  2  ±  V2~. 

112.  Turning  point  when  x  =  3  ;  points  of  inflection  when  x  =  0  or  3  ±VS. 

113.  Turning  points  when  x  =  k-jr  or  (2  k  +  1)-  ; 
points  of  inflection  when  x  =  (2  fc  +  1)  -  • 

V3  —  1 

114.  Turning  points  when  x  =  cos-1 ; 

2  2  7t 

points  of  inflection  when  x  =  ktr  or  2  for  ± 

3 

115.  Turning  point  when  x  —  ae  ;  .point  of  inflection  when  x  =  ae%. 

116.  12  ft.  1  119.  At  an  angle  tan-1  A'         120.  lain. 

117.  70°.  118,  V2'  ■   with  the  ground.  121.  5V5ft. 

Page  217 

122.  24  sq.  ft.  ;  14.71  sq.  ft.  per  second.         124.  2  V(5  -  «)  (a  -  3)  ;  4(4  -  s). 

123.  Greatest  distance  =13  ;  125.  a. 

force  =  —  (7  —  2s).  128.  -rr miles  per  minute. 

Page  218 

130    (b  sin  6  -\ — ^—  )  times  angular  velocity  of  AB,  where  0=  CAB. 

"  V  Va2  —  62  sin2  BJ 

131.  Circle. 


132.  kva2sin2&£  +  b2 cos2 kt;  maximum  at  end  of  minor  axis;  minimum  at 

end  of  major  axis. 

133.  2V2. 

134.  2  «w  sin  -  ;    w  Va2  —  2  ah  cos  <t>  +  h2,  where  w  is  the  constant  angular 

2 

velocity. 

,  __   2  aw  (a  +  &)   .    6    (a  +  b)  w    /-= — — . 

137. — ■ — -  sin  - —  Va2  —  2  ah  cos  9  +  h2. 

b  2  b 


ANSWERS  493 

138.  a0w,  where  a  is  radius  of  circle,  ad  is  distance  through  which  point  of 

string  in  contact  with  circle  has  moved  along  circle,  and  w  is  the 
constant  angular  velocity. 

139.  — ft.  per  second,  where  x  is  distance  of  man  from  center 
V10000  -  x2 

of  diameter. 

140.  ft.  per  second,  where  x  is  distance  of   man   from 
Vl0000-x2sin2a 

center  of  diameter. 

2  a2 
-  — times  man's  rate,  where  x  is  distance  of  man  from  center  pf 


a2  +  x: 
diameter. 

Page  219 

142.  Circle;     feet  per 

2Vb2-aH* 
second,  where  at  is  distance 
of  foot  of  ladder  from  side 
of  house. 

143.  tan0;^. 

a* 

Page  220 

155.  4  7ra,  4  ira  V2  ;  0,  8  ira. 

156.  w  Va2  +  2  ab  cos  8  +  b2,  where 

w  =  rate  of  6. 

157.  2,  V3;   },  1. 
165.  Greatest  when  x  =  (2  k  +  1)  -  ;    Least  when  X  =  few. 

Page  221 

168.  S  a  sin*  cos*.    169.  t-  *    170.  a(5-4^^)f,  171    2a2. 
b      a  9  —  6  cos  0  3  r 


144. 

—  tan  d> ;  — 

sec4  0  esc*. 

145 

ens/  _  sin/ 

cos£  +  sini' 

2 

e*(cos£  +  sint)3 

152. 

tan-1  |. 

153. 

0,*. 
2 

158. 

S(axy)i. 

162.  2  a. 

159, 

.'/- 

163.   ,;  \   Hi. 

161. 

a 

164.   J-V2. 

CHAPTEE  XII 
Page  253 

22.  2  Va2  -  x2. 
Page  254 

23.  x  -  log  (e*  +  1) .  30.  Bin*  3  x  (i  sin  3  x  -  ,2T  sin2  3  x 

24.  log[log(x  +Vx2-aa)].  +  TV  sin*3x). 

26.  Jsin*(2x  +  1).  3lJcos35_2cos?. 

27.  l-  (tan  ax  +  sec  ax).  ^    ^  3+2.sin24x    ^ 

28.  -  ^(ctnx  +  1)3.  6Vshil^  ' 

29.  ^(3cos52x-5cos32x).  33.  2sinx  -  log(secx  +  tanx). 


494  ANSWERS 

34.  §  sin8  x.  46.  -  4  eta  -  -  -  ctn3  -  -  -  ctn«  - . 
1                             %  4.     3         4      5         4 

35.  -(tan  ax  —  ctn  ax).  ._    .        .    .     ,   „,.   ,    ,     .    ., ,     .  „ 
a  v                           '  46.  log  sin (x  +  2)  +  \  ctn2  (x  +  2) 

36.  tan  2  X  +  sec  2  x  -  X.  -  \  ctn4  (x  +  2) . 

37.  -  5  ctn  3  x  —  ^  esc  3  x  —  x. 


47         1        „2x/_      „       „2x\ 
x\  *        10 


ctn»^(5  +  3cta2^y 

n3  ax tan  ax 

a 

)vcsc  ax  +  sin  ax) . 
a, 
„    9  3        x(      „x      _\ 
50.  -x  sec-  sec2-  — 7). 
x                                       7\       3\        3        / 


38.  2  (tan-  +  ctn    . 

\  '  48.  —  tan3  ax tan  ax  +  x. 

39.  £sin(2x  -  1)  -  ^sin3(2x-l)  3a  a 

+  fa  sin5  (2  x  -  1) .  49    _  1  fMB  r/j,  +  s}n  ^ _ 

/      x 

40.  3   sin — cos 


41.  -  tan2  -  +  3  log  cos 
2         3  3  51.  \x-  1LSin(6x  +  2). 

42.  J  tan  (3  x  +  2)  +  \  tan3  (3  x  +  2) .  52.   1  x  —  T^  sin  (4  -  6 x). 

43.  ?sec3^.  53.  x+lcos4x. 

9        _2 54.   Jgx-  J^sinl2x. 

44.  glj  Vtan  2  x  (7  tan  2  x  +  3  tan3  2  x) . 

Page  255 

55.  I x  +  J,  sin x  +  ^  sin 2 x.  72   1  s[n_l 3x _ 

56.  fe  x  -  ^4  sin  4  x  -  Js  sin3  2  x.  3  5 

57.  -  §  V(cos2x  +  l)3.  73.  ^tan-i22 


58.  log  (sec  x  +  tanx). 


6  V3 

1         ,2x 


59.  cosx  +  sinx.  74.  g"~~ 

cft    irsin(a-6)x      sin(a  +  ft)x~j  j  ,2x  +  l 

60>  ^  — T TTT —    "  75.  -  tan- 


2\_      a-b  a  +  0      J  ""e  3 


lTsin(a  +  6): 
2  L      a  +  ft 


y»(a-6)»1.  76>  _2_tan-i^±l 

a- ft       J  V7  V7 

62.  i-xsin6  —  |cos4x.  77_  ^n-i x ~  2 , 

63.  ^tCosOx— Jr  cos4x  — icos2x.  V6 

1     .     ,  4  x  —  5 

64.  I(tan2x  +  sec2x).  78.  -—sin-1 — 

v  2 

65.  —  log  (1+ cosx).  2x— 9 

x  79.  sin-1 — — — -■ 

66.  —  2  ctn x.  9 

80.  sec-^x  +  l). 

67.  x  +  ctn  x  —  esc  x.  7  x 

0  j  81.  logVx2+9  +  -tan-!-. 

68.  '-x  + (8  sin  2  ax  +  sin  4  ax).  o  * 

82.  ^log(3x2  +  2)--^tan 


2  V 2  cos  -  • 

2  83_  _Vl  — x2  — 2sin-1x. 


Vo  V& 


70.  log  (sec  x  +  tan  x) .  g4        in_  1 »  _  Va2  -  x2 . 
V2  a 

V2  ,  85.  -  tan-1  (cosx). 

71.  _(cos32x-3cos2x).  ^  ^log (2 x  +  Vi^T^) • 


ANSWERS 


495 


Page  256 


87.  £log(3x  +  V9x2-2). 

12      °3x  +  V2~ 

89.  Jlog(4x  +  3  +  2V4x2  +  Vx). 

90.  I  log ?LZ*. 

6     °     x 


91. 


1    ,      2x  +  3-V6 

—  log =• 

V  5        2  x  +  3  +  VS 

1 


~  log  (3 x  -  1  +  V9x2-6x  +  9) 

-log^+-7. 
10     °x-3 


94.  —  log(2x  +  2+V4x2  +  8x-14). 

V2 

95.  ^log(x2  +  6x  +  12)-GV3tan-i^i^. 
2  V3 

96.Ilog(xa  +  x-6)+llog^|. 

97.1log(2x2  +  5x  +  l)  +  ^log4X  +  5-V!!. 
2  34  4j.  +  5+Vl7 

98.  ilog(6x2  +  7x-3)  +  3log3-^i. 
2  2     62x  +  3 

99.  ]  log(2xa  +  6x+  9)  -  *  tan-i2x  +  3. 


100.  log (3x8  +  2x  +  8)  -IlJtan-i3g  +  1 

1  12  2V2 

101.  3sin-i^i_V3  +  2x-x8. 


102.  -V8-4x  — 4x*  +  7sin- 


103.  V4x-x2  +  uin-i  — 1 

104.  tan-i(x  +  tana). 

105.  sec  a  tan-1  (r-  sec  a  +  tan  a). 

106.  sin-1  (e*  cos  «  +  sin  or). 

107.  log  (x  +  Vxa  -  1 )  +  sec-1  x. 
1 


108.  -log(xa+Vx*-a*)- 


9 
110.          tan-1 

4tan-  +  8 

■\  7 

111.  — !— tim- 

4V2 

tan  x 

vT 

-2V2 


\  2       tan  2x  — 3  +  2V2 


3  tan  -  -  1 
109.  -log ! 

3  tan  -  +  1 

2 

Page  257 

pft  +  cxab  +  ex 

120. - 

c(l  +  log  a) 

121.  21og(e*  +  l)-x. 

122.  llog(e^-2)_l 


tan-  +  2  +  V5 

113.  —  log - 

Vs        tan-  +  2-V5 
2 


123.  log(e*  —  e~x). 

124.  l(x2  +  2x  +  6)j\_2T^3. 

125.  x  +  2  +  4  Vx  +  2  +  4  log  (Vx  +  2  -  l) 

126.  ^  (x2  -  2  a2)  Vx2  +  a2. 

127.  J  (x3  -  9)  v  x3  +  3. 


496 


ANSWEES 


V4 


141.  21og(2x  +  V4x2-9) 
V4x2-9 


129.  • 

27  V(9  -  x2)8 

130.  1^(3x2-8)(4  +  x2)f. 


142. 


143. 


ix2-  18 


6V3  +  2x2 

x2  +  18 


132.  Vx2  -  25  -  5 ! 


,__    V9x2 -4       9  ,3x 

133. sec-1 

8x2  16  2 

134.  !>,g(4x-l)(3x  +  l)3. 

x3_9x2_81x_8i 


135. 


+  271og(x+! 


2(x  +  3) 

136.  ^_i-log(3  +  4x3). 
12      16     GV  ' 

137.  -x(2x2-a2)Vo2^x2+— sin-i-. 
8    v  '  8  a 

x 
138. 


Vx2  +  9 

144.  -  x  (5  a2  -  2  x2)  V«2  -  x2 
8  ' 

3  a4.      ,  x 

+ sin-1-- 

8  a 

145.  ^(7x-5)(x  +  l)i 

T2  4-    2        , T 

146.  5-±j£  Vx^^l-Ssec-1-- 

x2  2 


Vx2  -  1 
1 
4  Vx2  +  4 


1 .      Vx2  +  4-2 
-log—  


140.  V3  -  x2  +  V3  log 


V3  -  V3  -  x2 


147.  _^liV2-x2. 

6x3 

148.  I  (x4  +  4)  Vx4  +  1 


1        Vx4+1-1 

+  -  log 

4         Vx4  +  1  +  1 

149.  J}  (2  x2  -  1)  Vl  +  4x2. 

150.  —tan-1- 

2  a  a      2  (a2  +  x2) 


age 

258 

151. 

3x2 
3(5- 

-10 

-x2)f 

152. 

sin-1 

2x 

(X- 

-3 

2)Ve 

153 

^log 

2     ° 

2x-l 

2x  + 

Vl6x2-12x 

+  3 

154. 

1   . 

-sm 
4 

2 

3- 

2x 

155. 

1     10, 

x  +  2 

V2     '  V2  x2  +  4  x  +  4 
156.  x  (log  ax  —  1). 


161.  xsec-12x-Uog(2x  +  V4x2-l). 

162.  -  sec-1 3  x  -  —  V9x2-1. 
2  18 

163.  517e3r(9x2-6x+  2). 

164.  l[(2x2-l)sin2x+2xcos2x]. 

165.  x[(logx)2-  21ogx  +  2]. 

166. sin  6  x cos  6  x. 

4       12  72 

167.  Je2x(2sinx  —  cosx). 

168.  TL  e?  (3  sin  3  x  +  cos  3  x) . 

169.  1  [xVx2 -  1  -  log(x  +  Vx2 -  l)]. 

170.  \  [sec  x  tan  x  +  log  (sec  x  +  tan  x)]. 


157.  — r(m  +  l)logx-l] 

(m  +  l)2"V 

158.  x  tan-1  ax log (1  4-  a2x2) .  171   g_  ■  \osx" ~~  *. 

2a  "2  &     x 

159.  x  log  (x  +  Vx2  +  a2)  -  Vx2  +  a2.  ^    g  ^  +  ^     x-2 

160.  £  (sin  3  x  -  3  x  cos  3  x).  &  x2  (x  4-  2)3 


ANSWERS 


497 


173.  lo£j 


x         .  1*        ix 

— ,  +  -  tan- 1 - 

Vx2  +  4      2  2 


174.  x  —  tan-!x  +  1( 

(3  +  1)8        3 


Vx2  +  1 


175.  log- 


xa 

4x  +  7 


Page  259 


181.  log  V<* +!>(»- 3)', 


x-2 

182.  log(x-2)(x2  +  2x  +  5)§ 
.,35  +  1 


+  -  tan- 
2  2 


183.  log 


x  —  4      x—  4 


184.    1f-8g+log/-£-Y. 

x(x-l)         °\x-l/ 


185.  log 


Vx^T 


</x*  +  X+l 

1    .        ,2x  +  l 
-tan-1 — 

V3  V3 


log 


Vx2  -  2  x  +  4 


2    '  x  +  2 

1C„     ,        "V/X2+l         1 

187.  log  +-t,an-i.i-. 

Vx  —  1       2 


!-l 


188. 


189. 


,     '        +ltan-ig. 

S(x2  +  4)      16  2 

1-x 


+  logVx2  +  l 


2(x2  +  l) 
+  \  tan-Jx. 

190.  ^logVI+^  +  Vg-2Vi+^. 

Vl  +  x-V2 

191.  x  -  1  4-  4  Vl-x 

-21og(Vl-x-x) 

6         2Vl-x  +  l-V5 


177.  lo< 


x  +  2 


1rfa    ■       Vx  -  1       l 

178.  log — ==== tan-ix. 

\  x-  +  1      ^ 


179.  log 
180. 

193. 


V(x-2)(x+4) 
x-1 
12x2  + 6x+ 9      3 


16xT¥-  +  4l0g(2*+1>- 


+  ^lo,V-c3  +  2-^ 


194. 


3Vj3+2      (JY2     ~~  \  x-+2  +  \  2 
1  +  sin  - 

log Bin  5/3  +  sin9  -Y 

,         .    x      3        2 1  2/ 

1  —  sin  -  x  ' 


195.  ±log1+COs4x        cos4x 


16        1  —  cos4x      8sin24x 
196.    1  [a  a  x2  +  «2 

-a2log(x  +  Vx2  +  u2)]. 
x 


197. 


198. 


v  1  +  x» 

vT+^ 


199.   kx(2x3  +  a2)Vx9~+  -/-' 


a* 


8 

^1  +  S 


log(x  +  Vx2  +  a2). 


203. 


}x(x2  +  10)Vx2  +  4 

+  01og(x+Vx2  +  4). 

3sin2x-2       3, 

7T~- 5~  +  ^l°g(secx  +  tanx). 

2  sin  x  cos2  x      2  ; 

COS  X  1 , 

-2sh^  +  2l0g(C8CS-Ctna!>" 


+  ^log 


V5      2Vl-x  +  l  +  V5 


192. 


2cos2x      4     °l  +  sinx 


204.  log  (sec  x  +  tan  x)  —  esc  x. 
205    sinx(2  +  3cos2x) 
8cos4x 
+  I  log  (sec  x  +  tanx). 


498  ANSWERS 

CHAPTER  XIII 

Page  285 

h       _  h 

1.  2.  2.  2.  3.  Ja2.  4.  a2(e°-e   <*).  5.  4  Tra2. 


Page  286 


7.  7r-2  1og2. 

8.  j}(e*  +  2).  15.  4a2. 
9-  f.  16.  8. 

10.  |(3 r- 2) a2.  17i  aWs. 


a-  +  a- 


h—Vw- 


20. 

J(4-7r)u'2. 

21. 

A(tt-2)«2. 

22. 

Tra2. 

23. 

a|V2. 

24. 

3  Tra2. 

11.  13J. 

12.  |a&. 

13.  2irab. 

18.  A (6 -a) 

15 

■     19.  ^aft. 

°V? 

—  a 

25.     gTTO2. 

Page  287 

26.  fmz&. 

27.  iTTrt. 

„„    2  a 
28. 

7T 

29.  .6366. 

30.  Jn2. 

2u2 

31.  — 1. 

Try 

32. 
34. 
35. 
37. 

§Trr3. 

§• 

+  7r3a2. 

2  a2. 

38. 

4n 

39.  *. 

n 

Page  288 

40.   i(4-7r) 

a*. 

41.   g  Tra2. 

42. 

11  TT. 

43.  le^a2. 

44.  a-V2. 
3 

45.  a2[V2- 

■log 

;(V2  +  i)]. 

46.  -  (3  tan—  +  tan3-)  c2,  where  2c   is  the  length  of  the  chord  through  the 

focus  perpendicular  to  the  axis  of  the  parabola. 

47.  J  to2.  49.  j(3V3  -  7r)a2.  51.   ^Tra2.  53.  fa2. 

48.  21i.  50.  ^(STr  +  9Vs)a2.  52.  2  tt  (a2  +  b2) .  54.  §  6a2. 


61.  —  (2a2+l)  +  2wa. 

4 

62.  §7ra3(l-  cos  a). 

M    9a3V3 

63. 

5120 

64.  §  Tra3  (3  log  2 -2). 

65.  Tra3  tan  6. 


Page  289 

55.  |a8tan0. 

57. 

|  a3  V§. 

56.  2Tr/i2Vp1p2,  where 
p1  and  p.2  are  the 

58. 

—  (3  a  — 
3    v 

respective  param- 

59. 

ffoira*. 

eters  of   the  pa- 
rabolas. 

60. 

38|tt. 

Page  290 

66.  VW«3- 

69. 

4Tr2a3. 

67.  4  Tra3  (2  log  2-1). 

70. 

%  ha2  {Sir 

68.  2Tra3(4-Tr). 

71. 

iha2. 

72.  2Tr2a6d. 

73.  1152  cu.  in. 

74.  22  ^tt. 


, 


AN  S  WEES 

499 

Page  291 

75.  *£-irV2.          76.  §7ra26.          77.  2n*a*.           78.  5Tr2a3.           79.   §  Tra3. 

80.  §(2  +  Tr)a3,  where  a  is  the  radius  of  the  sphere. 

81.  ^(l0Vi0-l).             83.  6a.                                  86.  -(log9-l). 

82.  |(e*-e~«).                     85.   |*ra.                               87.  8a. 

Page  292 

88.   §  o.                   89.  177.5  in.                     90.  log(e 

4-e-x).                   91.  i?. 

TT 

92.  ~[27rV47r2+l+log(27r  +  V47r2+l)]. 

93.  4<JA  2.                   95.  8  a.                   97.    ]  ;|  thi. 

94.  100.                        96.   §  a.                   98.  a  log-- 

h 

99.  2  Tra//. 
100.    ],2  Tra2. 

ioi.  v;  Tra-. 

Page  293 

!      Si               2  A               A                    A 

102.  2irah  +  —  (e«  _e     «  _  4e«  +  4e   "). 

103.  -^  Tra2.                         106.  4  Tra2  (2- V2). 

104.  «.-t7ra2.                                      „    ,- 
inR    ?     2                             107.  4Tra2V2. 

105.  4  Tra2. 

108.  2. 
a 

109.  9$  ft.-lb. 

110.   27!k\c-a~,  where  k  is  the  constant  ratio. 

111.       (2ar—  a-),  where  r  La  the  radiiu 
2  /•                   " 

i  of  the  earth. 

112.        ,  where  r  IS  tlie  radius  of  the 

r+  a 

earth. 

Page  294 

113.  2ttC.                    115.  1.11  ft.-lb. ;          116.1056io.            118.  58£ia. 

114.  189.6  ft.-lb.                  117  ft.-lb.               117.   12,',.              120.  168 w. 

Page  295 

121.  7f.                           126.  |  toaft2,  where  2 a  is 

122.  3041  §  lb.                                the  length  of  the 

123.  10,750ft.-lb.                       axis  in  the  surface 

124.  1250  tons.                             of  the  liquid. 

125.  10,416$ ft.-tons.       127.  fgirb. 

128.  l;U(87r+  9V3)w. 

129.  64  w. 

130.  L'i. 

131.  12.27  tons. 

Page  296 

132.  4.2  tons.                                                  136.  On  the  axis  of  the  segment,  | 

133.  i  7tm;.                                                                  of  the  distance  from  the  ver- 
/     2a\                                                             tex  to  the  base. 

(*-)"                         .sW4<"  *"\ 

\      5  /                                                  138.  Intersection  of  the  medians. 

500 


ANSWERS 


139.  ( — i  0),    where   x  =  h  is   the 

equation  of  the  ordinate. 

140.  (0,  |). 

141.  On  the  radius  perpendicular  to 

the  base  of  the  hemisphere,  £ 
of  the  distance  from  the  base. 

142.  y  =  A  &■ 


143.  Middle  point  of  the  axis. 

144.  On  the  axis  of  the  cone,  §  of 

the  distance  from  the  vertex 
to  the  base. 


145.     - 


146. 


Page  297 


147. 


2  a      a  (e4  +  4  e2  -  1)N 


148.  x  = 


fi  +  1 

2  a 


4e(e2-l) 


149.  (9,  9). 

150.  (|,  3). 

151.  y=-ik. 

152.  On  the  axis,  §  of  the  distance 

from  the  base  to  the  vertex. 


Page 
159. 
161. 

162. 
163. 


298 


\2    2(4tt-3V3) 

On  the  axis  of  the  solid,  TV  of 

the  distance  from  the  base  to 

the  highest  point. 

\2    2      lip) 

ha        aVs 
\T'  4 


153. 


/256  a 
\315tt 


256  a\ 

3157T/ 

'4_a     4(«  +  'j)n 


155.     0, 


3tt 
352 


5(2+  Sir)/ 

156.  (ira,  £a). 

157.  {ira,  |a). 

158.  x  =  -- 


164.  On  the  axis  of  the  lamina,  3f  ft. 

from  the  vertex. 

165.  On  the  diameter  of  the  sphere 

perpendicular  to  the  planes, 
and  half  way  between  them. 

166.  The  middle  point  of  the  axis  of 

the  solid. 


167. 


.=  !(VS 


■i). 


Page 
171. 
172. 
173. 

174. 

175. 


299 

M 


2M  L.     l  +  a 

log 

i2   \         a 

M 

5Ve! 

2cM/l  1 


— ) 

1  +  a) 


,  where  M 


I"    \e      Vc2  +  I2 
is  the  mass  of  the  wire. 

— ,  where  p  is  the  mass  of  a 
c 
unit  length  of  the  wire. 


176. 

177. 

178. 
179. 
180. 


SM 
25ir' 

Mc 

(c2  +  a'2)  * 
2cM/l 


\cM/l 


2M, 
a2l 
M 
60" 


Vc2  +  a2 


■a2-V(c  +  l)2  +  a2). 


ANSWERS 

Page  327 

CHAPTER  XIV 

24.  9z2  +  9(y-l)2  =  4(x- 

-i)-;  (i, : 

L,0). 

Page  328 

32.  (-2,  -1,5). 

33.  (3,-2,  1),  (-X.  |,  y 

■*■(*!,  tl.  A)- 

)• 

„   Vi       2V2       V2 

40.  , , 

6              3             6 

41.  2V3:V2:\  2. 

4 

35.  x2  +  ?/2  +  22-  2 x  +  4 ?/- 

■2z  =  43. 

42.  cos-i— — -. 

Vll4 

Page  329 

44.  cos- i 

53.  z  —  6  =  0,  4x-y-  5 

7 
46.  x  —  Sy  +  7z  +  39  =  0. 

54.   X  =  ^  =  ^. 

3             5 

47.  x  +  y  +  2  -  10  =  0. 

48.  x  -  z  +  1  =  0. 

49.  x  +  v  +  z  -  4  =  0. 

50.* 
3 

51.  cos-i^. 

55.  x-4  =  ^  =  i+J 
-5           3 

56.x-,  =  "  +  2  =  2_2. 
\  2 

58.    ft,  A,*|- 

Page  330 

,,        "             26             " 

67.  7  x  -  4  y  +  2  2  -  22  =  ' 

501 


V1086      V1086      Vl086  68-   X  ~  :'!'  +  ,J-  ±  4  N  ''-        " 


X  -.1       y  —  Bz  +  6 


69.   \  ::. 


2~_~1T~  =  ~7~"  70-  (0,l,2),(-§,  J,  |). 

a-i]].  *.  (L7S,  ... 

+  y  +  z  +  3  ±  3 Vs  =  0.  72>  1'"'  x  --• 


ez.  bid 

64.  x 

65.  4(/  +  9  =  0  and  2x  +  O2  -  1  =  0 

66.  (3,  -1,2). 


73.  :;./•  +  5// -z- 20  =  0. 

74.  7:4: -5. 


Page  331 

75.(3,0,-3).  82.  x  +  2y  +  z-2  =  0. 

78.  x  -  z  =  0.  83.  13x  -  y  -Viz-  32  =  0. 

80.  11  x  -  2y  —  bz  —  25  =  0.  84.  3 x  +  8  y  +  21  2  -  66  =  0. 

81.  19a:  +  10//  +  272-70  =  0. 

Page  332 

85.  ox  _y_z-5  =  0.  92.   (0,  1,  2),  (^P,  Jy^L,  -  ^J). 

86.  (-  1,  2,  -  1),  (2,  -  3,  -  4).  95.  y-  +  4  y  -  0  x  =  0. 

87.  3x  +  y  +  32  -  1  =  0,  96.  x2  -  z-  =  0. 

x  -  0  y  +  2  -  2  =  0.  97.  2  ./•'-  +  z*  -  Ox  =  0. 

88.  .V1..  98.  bi/2-2z2  +  8z  =  2. 

90.  (1.  I,  3).  100.   1/  =  x2  ;  2  =  x3  ;  22  =  tjs. 

91.  13x  +  11  y  -17z  -15  =  0. 

AC 


502  ANSWERS 


Page 

333 

102. 

xy  : 

=  1. 

108. 

COS" 

-i_L. 

V6 

Page 

334 

114.  2. 

115.^= 

2 

1 

110.  cos-i  — .  112.  e-e-i. 


3 
111.   |(12  +  5 log 5). 


113.  15. 


—  2 

——  =  Z  —  1;  2x  +  2y  +  z-7  =  0. 

v  —  1        z  /- 

116.  x  -  1  =  2 =  —7=  \  x-y  +  V2  z  =  0. 

—  1        V  2 

117.  ^—  =  2/  =  ^—  ;  4x  +  ?/  +  9z  -  39  =  0. 

4  9 

118.  4x  +  2iry  -  tt2  =  0,  y  -  z  =  0  ;  ttx  -  2  y  -  2z  +  2ir  =:  0. 

119.  K_l=^-=l_  =  ^ZLl;  x  +  32/  +  4z-8  =  0. 


CHAPTER  XV 


Page 

361 

1 

2/3 

X3 

(X2  +  2/2)f 

'  (x2  +  2/2)i 

2 

2/ 

X 

X2  +  y2 

'  x2  +  i/2 

3. 

2/ 

1 

x  V2  xy  — 

y2         V2  xy 

-y- 

4. 

X 

"VV  -  ; 

*2(2/+V2/2- 

X2) 

1 

V?/-; 

c2 

5 

2/2 

xy 

(x  -  yf 

x  —  y 

X*       co°    X2/ 

(x  -  y)2 

x-y 

Page 

362 

\2/  x/     y2  y 


15. 


1+2/     (1  +  2/)2 
2/  2/ 


x^  +  y2     x2  + 


17. 


2/ 


&-v2       xVx^" 

2y     2  \ly 
19.  2  tan-2'         2a" 


x      X2  +  2/2 


20.  2  e^sin  (x  -  y).  30.  .640  sq.  in. ;  .650  sq.  in. 

27.  .0004.  31.  .49  ft. 

28.  -.0302;  .0002.  32.  .0213ft. 

29.  11.030  cu.  ft.  ;  10.996  cu.  ft. 


Page 

363 

33. 

.0048. 

34. 

1 
Vx2  +  y2 

;  o. 

35. 

0;  0. 
1-V3 

36. 

2 

Page 

364 

k 

41. 

y 

42. 

x  +  y 

X 

43. 

y      Vx2- 

>r 

X 

xsin- 

it 

ANSWERS  503 


37.  4x  +  3t/-11  =  0. 


38.  k  v  z{  +  y2  in  direction  tan-1  -3 . 

1  V3 

39.  _-cos0-  — -  sin^;  1. 


45  Sx*z  •  z(1-2v2) 

'  2  -  2  z2  '  ?/  (2  z2  -  2) ' 

46  2aaJ  +  3zV  2yz  +  3x8y2 
3z2  +  x2  +  y2'        3z2  +  x2  +  2/2 

47  0j/z-2x(x2  +  y24-z2)2- 
"  2  z  (xa  +  ?/2  +  z2)2  -  9  xy  ' 

9xz-2j/(x2  +  y2  +  g2)a 
^         (y  +  z)(2x  +  y  +  z)  .  2z(x2  +  y2  +  z2)2 -'.»// 

(x  +  y)(x  +  y+2z)'  48.  2x  +  3i/+ 2z-9  =  0; 

_  (z  +  x)(x  +  2y  +  2)  x-2  _  y-1  _  z-1 

(x  +  2/)(*  +  2/  +  2z)'  2      ::      3     "     2 

49.  z  +  2y  +  z  —  2  =  0;  x-l=V  =  z-l. 

50.  2(ax.  +  by.)  (ax  +  /»/)  =  z  +  z,;  ^^ = V  ~  ?/* =  Lzfl. 

1    2a(ax1  +  6!/1)      2  6(ax1  +  6y1)        -1 

54.  z  +  y  +  z  —  8  =  0,  X  —  2  y  +  1  =  0  ;  2 x  +  //  -  3 z  =  0. 

55.  x  —  1  =  0,  2  //  +  2  z  —  tt  =  0  ;  2  y  —  2  z  +  tt  =  0. 

56.  4x  +  3y-26  =  0,  3y-4z-25  =  0;  3x  -  iy  -  3z  -  12  =  0. 


Page  365 

57.  4x-  Ay  +  z  -  24  =  0, 
2x  +  8?/  +  5z  =  0; 
14x  +  9y—  20z  +  79  = 


ieVr2-a? 

5H. 

sin-1 

rVa?  +  k* 

59. 

TV 
2' 

60. 

0. 

61. 

10\/2 

,  10^2,  5V2. 

a    a 

a 

62. 

3'  3* 

3 

63. 

Mean 

point  of  the  vertices. 

64. 

8c*c 

3V§ 

65. 

(-  u 

_  1  •"   a  "\ 

4  B>   4:1' 

66. 

('- 

1         5V2\ 

67. 
68. 

(-1, 

8abh 
27 

-1,1). 

69. 

2x  +  2y  +  z  —  6  = 

70 

x  =  — 
a- 

2aZ 

!  +  62  +  c2 ' 

y  =  - 

2  67T 

!  +  &2  +  c2 ' 

2cK 

a-  +  B2  +  c- 


74.  -F-  +  x. 


504  ANSWERS 

Page  366 

71.  x5  -  x3y  +  x2y2  +  y&.  76.  Vx2  +  y2  -  y. 

72.  xy  +  logxy.  77.  \  (x  +  sin  x  cosx)  +  ?/  cosx. 

73.  logx  -  A  _  7g    _  eTcos  (x  +  y)t 

79.  2i;2f. 

80.  1;  1(4  -7T). 

81.  If;  1^. 

75.   _e    ?-  logy.  82.  l  +  log2;  £(l+log2). 

Page  367 

83.  tan-H;  rfn-l|.  8?    ^      ^  F,(  ^/ [2^)]2 

84.  §7ra2.  2x2  T     3x3*/      V  '      dy2 L     V  'J 

85.  —(36- 2a).  +  &F"(x). 

b  dy 

im2  d2V      1  dV 

86™  (36  + 2a).  95       "+-J1. 

b  dr2       r  dr 


Page  368 

/a2F     d2v\  ,  .  .      „.  a2F 


h 98.  x?/( — )+(x2  —  y2) y hx 

du2      dv2  \dy2       ax2/      v         "  'dxdy         dx         dy 


CHAPTER  XVI 
Page  394 

1.  6 -log  2.  7.  iTra2.  11.  2|. 

2.  £(2-tt).  8.  2f.  12.  1(15 -8  log  4). 

3.  ir-2.  *  I0'-  13.  «(a  +  6)Va6. 

4.  1(11 -16  log  2).  10.^(44  +  9^).  14* 

5.  1(5  +  6  V3).  30 V  ;  T5 

15.  4  +  ^(Sin-i|-sin-iJ).  n<  ^(gr_2). 

16.  (  2  -  7T  +  5  sin-1  —  )  a2  ; 

V  V5/  18.  ^(3V3  +  2tt). 

/67r-2-5sin-i-Ma2- 

V  V5/  19- 

Page  395 

20.  10  7T.  24.  317.  29.  j^-n-a*. 

2  25    A2-  30-  i|(a'^  +  a62)Va6. 

21.  |(8  +  ir).  105'  1024 V2 

26.  -—  (e*™-l)(e2™-l).  '       315 

22.  -(tt  +  3-3V3).  10"  32.  ^ira*. 
6                                 27'  W-  33.  §ira*. 

23.  22^.                              28.  {TTtt4.  34.  r%%ah3. 


48 
*2V3 


answ: 

ERS                                       505 

Page  396 

35.^(15^32). 
a4 

a4 

38.  —  (15  tt 

24 v 

a4 

-32).                   «•<¥»»■ 
12.  (V-,  0). 

36.  -(8  +  3tt). 

39.  (45tt-128).                      /o9„      v 

1920                                   43.    (^,0\. 

37.  UiraK 

40.  T%7ra4. 

\15tt      / 

11   I      18a 

18  a      \ 

46    /     8a(6,+  8)x 
V       5(6w-4)  / 

\5(3ir-2)'  5(3 tt- 2)/ 

«(•?..). 

..    *    .,         .      .„.                  3  a  sin  a 

17    \'n  the  avis  ui  the  cector 

'      4  a 

from  the  center  of  the  circle. 

Page  397 

48.  ( , 

\105(16-3tt) 

304  a        \ 

53./a<37r+2>,0\ 
V      6V3           / 

105  (10-  3  tt)/ 

•^■•> 

51.  (            "■        ,0V 
V     2(a6-r2)      / 

50.  (ff,  0). 

55.  ^ttu-. 

56.  4ttV5. 

-■(-35-* 

57.  8  a2. 

58.  2a2(7r-2). 

*(-.). 

59.  ^(tt-2). 

3   V           ' 

60.  8  a2. 

Page  398 

a2 
61.  —  (20-3tt). 

68.    ,',;  rra4. 

18V                ' 

69.  -  \((P-Se  +  1B). 

62.  $(20- Sir). 

70 . 

2  a2 
63.  (20-3tt). 

"  36       48 

9   v 

71.  a(2-V§). 

64.  4a2[V2-logi 

Ci+Vi)]. 

72.    i(4  +  7r)-log(V2  +  l). 

65.  16a2[7r-V2- 

-log(l  +  V2)]. 

,«    «- 

66.  (71- -2)  a2. 

73.  — 
40 

67.    -(7T- 

8V 

•2). 

71.   27ra2. 

Page  399 

90 

76.  4,1,,. 

77.  §  ttu3. 

80. 
81. 
82. 

102$. 

Ott" 

7ra3. 

87.  \ira?b. 

O/78 

88.  —  (3tt-4), 

„„     a3 

89. 

360 

78.^. 
32  6 

83. 
81. 

9  n3 

5  a  . 

3  2«8 

-ij-a  . 

»*(. 

Vs- 

-4). 

85. 
86. 

§  7ra6c. 
X7ra3. 

506  ANSWERS 

Page  400 


91    —  (3 ir  +  20  —  10  V2)  96'  §to<i'  where  k  is  the  coefficient 

9  of  variation. 

92.  -  (17 -  12  log  4).  97.  ^-  (a2  +  4  ft*). 


Atto8. 


98.  2917T/0. 

99.  $Mh*. 


94.  §7ra3.  ^   3M(a!-af\ 

Page  401 


10  It 


10L^/a2-  107.   i^^al  +  S^Xol-^i 

102.-(2/i2  +  a2).  %«-  +  8*K.?-*)*,. 


8  a      152  a      11  a\ 
9    '  """  \T5'      525  '     120/ 


103.  ^(3tt-7).  108.  Z8" 

\15 


104.  ™£(r| -*•«).  109    /asin^    a^-cos^)     kO, 

15  '  Wi      '  <?i  '    2 

105.  —  (4  62  +  3  a2) .  ^  =  2  mir. 

,4    „  ,„    /3  a      3  a      3a\ 


Page  402 

/8a      3/)      3c\  117.  At  the  middle  point  of  the  axis 

"  \g~ '    ~g~ '    "§/  '  of  the  cone. 

u,  (0,  »  $.  -  fv  T-  t); 

n,  (0,  0,  |). 


119.  On  the  axis  of  the  cone,  |  of 
the  distance  from  the  vertex 
to  the  base. 

120.  On  the  axis  of  the  shell,  21  §1 
3  -xb  sin  a  from  the  center  of  the  spheri- 


115"       16  a      f r°m  °enter  °f  Spbere"  cal  surfaces 


116.  On  the  axis  of  the  hemisphere, 


121.  On  the   axis   of   the  frustum, 
ft(r2  +  2r1r2  +  3r|) 


from  the 


15 

base.  smaller  base. 


ANSWERS 


507 


from  the 


Page  403 
122.  On  the  axis  of  the  cone 
q(407r  +  27V3) 
40~7r 

vertex. 

3  a2 
124.  2ttP. 


123. 


125. 


irpa 


(2  b -a). 


— — -  (36  —  2  a),  where  k  is  the 

coefficient  of  the  variation  of 
the  density. 

2_7T/J 

5 
4  3/(7- 4  V2~) 
15  a2 


Page  404 

130.  ^(2  +  V2). 


132. 


4V2-5V5  +  VT7  + 


131. 


3  3/(1 -cos  a) 
2  («2  +  a^o  +  a!;) 


^(I  +  VS)20 
(1  +  V2)4(l  +  Vi7)16j 


Page  433 

17.  -l<x<l. 

18.  -l<x<l. 

Page  434 


CHAPTER  XVII 

19.  -1<x<1. 

20.  -  2  <  x  <  2. 


21. 


-  3  <  x  <  3. 

I  l 

7=<X<  -^- 

V2  V2 


2        2-4        2-  4.  <; 

29.  x+  \x*+  ,2.x*  +  :i\':,x'+  .... 

„„    ,      x2      5x4      61  x" 

30.  I  4-  —  + + +   •  •  •  • 

\1       li         16 

31.  1  + j  +  x-+  |x8+  ■■•. 

32.x-1.^+1^-^-1:^^  + 
2     3       2-4     5       2-4-6     7 


■J  „  *" 


-K)-¥<¥*4 


34.  

V2 


35.1og5+^^-^^^+1(X-5)3 


2         52 


3         53 


3^[1  +  (-4)+(^  +  fe^+...]. 

7T         j_l  (X-1)2         (X-l)3 

37.  -  + -5 -  + +  •■.. 

4  2  4  12 

38.  V5[l+  §  (x  -  2)  +  5-V(x  -  2)2  -  Ti5(x  -2)8  +  ■•  ■]• 


5.08  ANSWERS 

39.  .2070.  42.   .8480.  45.  .00315;  1.0980. 

40.  .7103.  43.   -.0502.  46.   .22314;  1.0004. 

41.  .0848.  44.   .40546.  47.   .8473;  1.040. 

Page  435 

51.  2.0305.  52.  2.9625.  57.  |  R  |  <  .0004  ;  7. 

Page  436 

62.  1.328;  1.300;     64.  -.0062;  -.0050;      „w    .      a 

67.  loe- 
1.308.  -.0991.  &6 

63.  .14;  .1325;         65.  -  2  V3.  68.  0. 
.1418.                   66.  3.  69.  2. 

Page  437 

78.  |.  81.  0.  84.  1. 

79.  6.  82.  -  §.  85.  1. 

80.  0.  83.  1.  86.  e.  88.  1. 

7T2  .  /COS  X         COS  2  X         COS  3  X  \ 

9i-  j-i-v"—+^ — )• 

eaK  —  e~  air      a  (e""  —  er  an)  I  cos  x         cos  2  x  '  cos  3  x 

*         2^  w  \l2+a2_22  +  a2  32  +  a2 ' 

gf<77  _,.  e-  on-  /  s;n  x        2  sin  2  x      3  sin  3  x  \ 

+  


60.  \B 

l< 

.00005. 

70.  -  J. 

74.  0. 

71.  -  ^ 

75.  0. 

72.   },. 

76.  0. 

73.  J. 

1 
e 

77.  2. 

89.  e. 

90.  1. 

/sin  x      si 


\12  +  a2       22  +  a2 

n  3  x      sin  5  x  \ 

__  +  ___  +  . . .  j  . 


_.    7r      2  /cosx      cos3x      cos5x  \       /sinx      sin2x  .  sin3x 

37r      2 /cosx      cos3x      cosSx  \ 


-•) 


sin2x      3  sin  3  x      sin4x 


/3sin 


7T2         „  /cos  X        COS  2  X        COS  3  X 

2  1 -I •••)  +  -     I 1  sin  x sin  2  x 

0  \   l2  22     '         32  /      7T I  \  1        I3/  2 


,     /7T2  4\     ...Q  7T2     .      ,         , 

+  ( )  sin  3  x sin  4  x  + 

\  3       33/  4 


CHAPTER  XVIII 
Pag3  471 

1.  x  +  y  +  log(x  —  l)(y-  l)  =  c.  7.  x2  +  2j?/  +  2/2-10x-6?/  +  c  =  0. 

2.  4  tan  x  +  2 y  +  sin 2y  =  c.  x2      c 

3.  7/(x  +  ?/)  =  ex.  8-  *  =  3-  +  "  +  L 

4.  \  x2  +  y"  +  xlogx  =  ex.  e3X 

.v  9.  ?/  =  —  (3  x2  -  x)  +  ex. 

5.  x(e;c  +  1)  =  c.  0 

6.  y-  +  0  xy  +  2  x2  +  2  y  -  8  x  =  c.  10.  x  -  y  -  1  +  ce~  'J. 


11.  y 


12.  y 


13.  y  = 


x  +  1       (x  +  l)2 

c  —  (x  —  lje* 
V3(l  +  x) 


V2(c-x3) 
Page  472 

18.  log (z2  +  y2)  =  2  as  +  c. 

19.  sin- !  -  =  log  x  +  c. 


»  +  —  =  c. 
2 


14. 

x2  +  cP"J  4-  sir 

i2/  = 

c. 

15. 

16. 

x3  +  2/3 
y  =  ex. 

1 

X 

1/ 

x2</2 

X2 

17. 

log(x- 

z/)  + 

tan- 

2/ 

:  C. 

509 


21.  ?/2  +  log  cos3  (x  +  y)  =  c. 

22.  sin  x  —  x  cos  x  —  tan  y  —  c. 

23.  x  =  sini/  +  ccos?/. 

24.  x2  =  c2  -  2  cy. 

25.  x2  +  i/3  =  ex. 

26.  0/  +  2x-l)2  =  c(//-x-l). 

27.  j/  =  ex. 

28.  x/(x-l)  =  (x  +  l)[x  +  c 

-log(x4-l)2]. 

29.  sin-  *  x  —  x  Vl  —  x2  +  2/  Vl  —  y- 

—  sin-1?/  =  c. 

30.  v/4  =  c(2/2+2x2). 

31.  tan-1?- 


34.  lo- 
36. 


2/-3X 


2x 
+  — =  c. 


</       Vx2  +  j/2 

32.  (x24-2/2)e*  =  c. 

33.  y  =  (x2  +  c)  e*3. 

Page  473 


47.  x-.v-'  +  2  e'  =  e.    

48.  y  =  x2  —  1  +  c  Vx2  —  1. 

49.  2xi/  +  log^  =  c. 

x8 

50.  y  =  x  (logx  +  Ct)  +  c2. 

51.  ?/  =  x  sin-1  x  4-  Vl  —  x2  4-  c:x  4- 

52.  y  =  CjX  4-  ca  —  log(l  4-  x). 

53.  2/  =  c.x  +  c., ;  log  cos  ax. 

a2 

54.  2/  =  er  (x  —  1)  4-  c^x-  4-  c2. 

x2 

55.  y  —  c,  lo?x f-  c„. 

4 
2 

56.  4  y  4-  c„  =  c.x2 log  x. 

57.  y  —  cx  secc1(x  —  c2). 


log  Vl  4-  x2  4-  c 
x2  =  c2  +  2  c?y. 
2/  Vsin  2  x  =  cos  2  x  +  c. 
x  4-  .'/  4-  Vx2  —  y2  —  c. 

J  1 

2~ 


■+*-=7+n 


2/  =  x2  -  2  4-  ce 
^tan-^  +  log 


3  ~  Vx2  4-  2/2 

Z  4-  r,  ,,  _  log  (10  x  -  15j/  +  7)  =  C. 


logVx2  4-  </2-  2  tan-' 


x  +  tan-1  —  =  c. 

r 


46.  2/ 


'l  4-  x2  Vc  -  x2  +  log  (1  +  x2) 


58.  y  4-  '',  logl/  =  x  +  ftg. 

59.  ,/  =  <•„  4-  Cj  sin  x  —  x  —  h  sin  2  x. 

60.  a  =  Cj  sinA:(x  —  c2). 

61.  2/  =  e2  4-  Cje-  *  —  \  (cos  x  4-  sin  x). 

62.  y (eV<-c2>  4-  e-^^-V). 

2  c, 


$.    y=    A(eA(x- 


e-A(..-c2)). 

3x 


64.  0  y  4-  c2  =  qx3 

65.  1296  2/ =  (x  +  l)4- 

66.  y  =  Vi  tan  V§  (x  -  ^J 

7T 

67.  sin  ?/  =  x  —  -  • 

68.  x  =  log  (V2  sin  y). 


510  ANSWERS 

Page  474 

69.  FSBVM+ V-i-+£  +  *-L  n-  ?'  =  V+ V^"+*«-- 

70.  y  =  cxe*  +  c2e- ■■        2  72-  ■"  =  (cix  +  h) e~**+{x-  2)2. 

—  -}-  (0  sin  2  x  +  2  cos  2  x).  73.  y  =  c2  cos  3  x  +  c2  sin  3  x  +  J  e3*. 

74.  ?/  =  e2  (cj  cos'— 1-  c2  sin— — )  +  x3  +  3x2. 

75.  .v  =  (x  +  c^ t-3r  +  c2e-6*. 

76.  y  =  c1  +  c.,e'Sx  +  x'2. 

77.  y  =  c,c-"--,:  +  c2e2x  —  —  (sin 2a  +  2  cos2x). 

78.  y  =  e~2x  (Cj  cos x  +  c2  sin x)  +  J  (3  sin  3 x  —  cos  3  x). 

a-  8x 

79.  y  =  cxe   2  +  c2e  2  +  0^  (8  cos  x  —  14  sin  x)  —  ?£F  (8  sin  2  x  +  19  cos  2  x) . 

80.  y  —  eP° (c,  cos  x  V2  +  c„  sin  x  V2 )  H (5  cos x  —  4  sin  x). 

41 

81.  y  =  (c,x  +  c0)e--2*  +  (0  -  20 x  +  25  x2). 

(525 

82.  y  =  C-pSinSx  +  (c2 Jcos3x. 

83.  y  =  qe2*  +  c.2e~x  —  — (3  sinx  +  cos  x). 

(x      x2\                      1 
Cj 1  +  c2e5  x  H (9  sin  x  +  7  cos  x) . 

85.  y  =  e-*(c1cosxV3+  e2sinxV3)  +  J(2x  — 1  +  e2'). 

xex'       1 

86.  y  =  c1e3j'  +  c„e-a: (4  sin2x  +  7cos2x). 

4        65 

87.  y  =  e2x(c1cos2x  +  c2  sin2x)  +  J,  (1  +  4x  +  4x2)  +  ^  (3  cos  3 x  +  30  sin  3 x). 

88.  y  =  c1e~x  +  e2(  c2  cos +  c3  sin )  —  0  +  x  +  x3. 

89.  y  =  c-je-1"  +  c2e~ x  +  c3e-'ix  —  J  (20  —  Ox  +  9 x2)  +  }  (cos 3 x  —  sin  3 x). 

90.  y  =  Cle-*s  +  e**(c2  +  c3x  -  ^  +  ^  +  J  (1  +  x  +  x2). 

Page  475 

9l.J,  =  ,(,,_|)+e,.^+?)+,,c-=J  +  ^. 

92.  y  =  cx&  +  U2  -  ^J  cos  2  x  +  (cs  -  |Asin  2  x. 

1  e2x 

93.  y  =  Cjex  +  c2e~ *  +  c3e- 2a'  +  -  (cos  2  x  —  sin  2  x)  +  —  (12  x  —  19). 

gX 

94.  y  =  e~x(c1  +  c„x  +  c3x2)  +  —  (2  sinx  —  11  cosx). 

95.  y  =  (cx  +  c2x)  cosx  +  (c3  +  c4x)  sinx  —  4  +  x  +  x2. 


96.  ?/  =  e2x(< 


i  +  C2X  +  t)  +  e_2x(cs  +  C4X)- 


ANSWEES 


511 


97.  y  =  c1e2x  +  c2e~2x  +  cs  cos  3 x  +  cA  sin  3 x  —  - —  (sin  3 x  +  3  cos  3 x). 

98.  y  =  e, +  c„x  +  c3  cos  xVi  +  c4sinx  V2H (13  sin  2  x  +  16  cos  2  x). 

6        17 

99.  y  =  c1  +  c2x  +  c3x2  +  cie~ix+  T2-^(cos4x  —  sin4x). 

100.  x  =  (Cjt  +  c2)  e<  +  e2 ',  </  =  (Cjt  +  c2)  e'  +  r3  +  e3 '. 

101.  x  =  c1sini  +  e2cos£  +  |  (e*  +  e-'),  y  =  c2sin£  —  CjCost  +  J(t'  +  e-'). 

102.  x  =  c1  +  ce'  +  c3e-',  y  =  c4  —  c2e*  +  cge-'  —  cos£. 

103.  x  =  cxe'  +  c2e5'  —  \,y  —  3cxe'  —  c2e6<  -  2. 

104.  x  =  c1t-2'  +  c2e2'  +  -Ie3<  +  ["g  (7  sin  3 1  +  3  cos  3 1), 

c  3  1 

?y  =  ^e-  2 «  _  _?  e2<  -  -  e3<  +  —  (sin  3  f  -  6  c< *  3  /)• 
3  o  13 

105.  x  =  c1  +  c2e«*  +  c8e-»'— Tfo  (49 1—  83«a  +  6*8), 

2/  =  2 e,  -  H  +  ?£ e2<  -  c3e-3'  +  1  (11  f -  3«»  -  6 t3). 
54        6  64 

106.  x  =  c,  +  c.,e-2'  +  c8e»<  +  |  (3*a  -  0, 

y  =  c,  -  ^-Sc^'+eV-  }(8P  +  St)< 

,„„  '"  /  at  .      at\        --%/  nt  .     at  \ 

107.  x  =  evi    c,  cos — =  +  C-Sin— —  )  +  e    va/c.cos — =  +  f ■  .sin      _  ). 

V1        -y/2  V2/  V  V2  V2/ 

at  a« 

77;/  "<  •      "'  \   ,     ~    '.,/  "'  «t  \ 

?/ =  eV2(  c„cos — -  —  CjSin — =)+e    va/c8sin        —  c.cos    —  )• 

V  "        V2  V2/  V         V2  V2/ 

Page  476 

ic8.  „  =  c.^/i  + ■ . ?'  +  L« . **  + . . .  +  ;•«•«- 6  +  S . 5E  + . . ) 

1     V        7     [1        7-8      [2  7  •  8  • » •  •  •  (r  +  G)      Jr  / 

+  ^(15  +  18  x  +  9xs  +  2x8). 

x3 

109.  y  =  Cj  (35  -  42  x  +  21  x2  -  4  J-3)  +  ^  (3  -  14  x  +  21  x2). 


XG  + 


6    [1      6-7    [2      6-7-8    [3 
+  4-5-6.  -(r  +  3).g  X 

v       ;   6  •  7  •  8  •  •  •  (r  +  5)    [r  / 

/,      n    .,      »(n-4)    .      »(n-4)(n— 16) 

ni.y  =  c1(i  +  i*  +  ^r-ix*  +  -   -f 

n(n-4)(n-16)...[>-(2r-2)g]  \ 

III  "7 

+  C2X(1  +  !Lzlx2+(»-1H--^x4  +  ("-i)(n-7Q)(n-25)x6  + 


(n  -  1)  („  -  9)  (n  -  25)  ■  ■  •  [n  -  (2  r  -  l)2]  ^r 


1 2  r  +  1 


^+...). 


512  ANSWERS 

112.  y  =  cJ\  +~  +  „    fr    n  +  • 
\        2-5      2-4.5-9 


[2  -  4  -  6  •  •  •  2  r]  [5  -  9  - 13  -  -  •  (4  r  +  1)] 


+  c0£i( 1  + 


2-3      2- 4-3-7 


[2-4-G...2r][3-7-ll.--(4r-l)] 


113.  w  =  Cl(l-  — +  — + 


+  (-  l)1- 


22r|2r_+l 


'\        22|2      2*14 


T4  r 


115. 

x  =  log(y  +  3). 

118. 

r{0  +  c)  =  l. 

116. 

k  (x  —  1)  y  —  y  +  1  =  0.    where 

119. 

r2  =  2(0  +  c). 

A;  is  the  constant  ratio. 

120. 

rn  =  c  sin  nd. 

117. 

y  =  ex". 

121. 

x'2  -  y2  =  c. 

Page 

477 

123. 
124. 

x'2  +  y'2  =  ex. 

a    I+f        _^±£. 
y  =  -(ea+e      «). 

129. 

y  =  ce   k  ,  where  x  =  a  is  the 
fixed  ordinate,  and  &  is  the 
constant  ratio. 

125. 

y  =  ^(ek(J-c)+e-kCr-c)h 

130. 

y  =  -(e   A"    +  e      *   ),  where  fc 

where  k  is  the  constant  ratio. 

2 
is  the  constant  ratio. 

126. 

.      Va2  —  2  ay  —  a 
x  +  c  =  4-  a  log — 

131. 

2/  =  c  ±  1  Vfc2x  —  4  x2 

k2  .      .  2  Vx       . 

±  —sin-1 ,  where*  is 

4               A; 

Va2  —  2  ay  +  a 

±  2  Va2  -  2  ay. 

127. 

r  =  ce        *■ 

the  constant  ratio. 

128. 

y  =  ax2  +  6. 

132. 

s  =  25  (2)  s  . 

Page  478 


134. 

p  =  14.7e-,00004A, 

135. 

$1218. 

136. 

67  niin 

137. 

EIy  = 

/lx2 

—  wl 

\2 

x3\ 

138. 

EIy  = 

w/l2x2 
2\~2~~ 

-?* 

-) 

12/ 

139. 

EIy  = 

w/lx2      X 

2\T~"e 

:)• 

133.  c  =  .01  e- -183* 


140.  (x  -  rx)2  +  (y  -  c2)2  =  c2 


141.  Cl2/2-^-(x  +  c2)2  =  l. 
142. 


V3^ 


ANSWERS  513 

Page  479 

-..*    a    ittt     i_        ,  .    .,  149.  About  7  mi.  per  second. 

145.  -  V3A:,  where  A;  is  the  constant  „       ,- — -            * 
2  150.  V2gh. 

ratio.  151.  20  sec. 

146.  J (4-  V2)  Vol  152.  234|  ft. ;  4.3  sec.  (g  =  32.) 

147.  ■ 153. 


J.  »1X/  mg  ,. 

V  mg  +  fcoj 


vi  + 

Page  480 

154.  1.8  sec.  (£/  =  32.) 

155.  s  =  c,  cosht  +  cn  sin  /i(  4 — cosfcf,  where  h-  is  the  constant  ratio 

K1  —  fca 

s  =  Cj  cos  Af  +  c2  sin  ht  H sin  kt,  if  h  =  k. 

It         t^P-ih*  t^/p-th- 

156.  s  =  e~  2  (qe       2        +  c2e  a       ) 

,  a(/t2  -  k2)  coskt  +  akl  sin  kt    ..  ,      „ , 

+ (tf-*y+(tt)' '  lf  *>2*; 

-£/  «V4/i2-  J2  .    fV4A2-/2\ 

2  I  c,  cos |-c„sin — I 

\  x  2  2  2  / 

a  (A2-  fc2)cosfe<4-  akl  sin  W 
+  (A2-Jfc2)2+(tt)2  '**<2A; 

Vl       2'  (A* -*»)*  + (2  M:)3 

(/(-  and  /  arc  the  constant  ratios.) 

157.  h  =  fc,  I  very  small. 


s  =  e 


INDEX 


(The  numbers  refer  to  the  pages) 


Abscissa,  5 

Acceleration,  178 

Angle,  59,  166,  205,  315,  316 

eccentric,  108 

vectorial,  118 
Arc,  differential  of,  174,  206,  324 

limit  of  ratio  to  chord,  172 
Archimedes,  spiral  of,  120 
Area,  of  any  surface,  381 

as  double  integral,  376 

as  line  integral,  354 

of  plane  curve,  143,  262,  267 

of  surface  of  revolution,  274 
Asymptote,  26,  79 
Attraction,  283,  393 
Axis  of  conic,  76,  80,  81 

Bemouilli's  equation,  446 
Bessel's  equation,  470 
Bessel's  functions,  471 
Bisection  of  straight  line,  9 

Cardioid,  124 

Catenary,  54,  441 

Center  of  gravity,  278,  379,  392 

Center  of  pressure,  277 

Change  of  coordinates,  39,  124,  386, 

387,  388 
Circle,  70,  108,  122 

auxiliary,  108 

involute  of,  112 
Cissoid,  85,  124 
Comparison  test,  406 
Components,  of  a  straight  line,  312 

of  velocity,  180 
Concavity  of  plane  curve,  167 
Cone,  306 


Conic,  83,  125 

Constant  of  integration,  147,  222 
Continuity,  136,  339 
Convergence,  405,  409 
Coordinates,  Cartesian,  1,  4,  301 

cylindrical,  385 

oblique,  43,  301 

polar,  118,  386 
Curvature,  207 
Curves,  in  space,  322 

intersection  of,  29 
Cusp,  109 
Cycloid,  109 
Cylinder,  303 

Degree  of  plane  curve,  29 
Derivative,  136 

directional,  343 

higher,  162,  338 

in  parametric  representation,  204 

partial,  335 

sign  of,  138,  166 

total,  344 
Differential,  141,  339,  359 

of  arc,  174,  206,  324 

of  area,  146 

exact,  349 

total,  341 
Differentiation,  136 

of  algebraic  functions,  154 

of  composite  functions,  357 

of  implicit  functions,  162 

partial,  335 

of  polynomial,  137 

successive,  162,  338 

of  transcendental  functions,  192 
Direction  in  polar  coordinates,  205 


',11 


INDEX 


U5 


Direction  cosines,  314,  319,  324 

Directrix  of  conic,  80,  83 

Discontinuity,  finite,  427 

Distance,  of  point  from  plane,  320 
of  point  from  straight  line,  63 
between  two  points,  5,  313 

Divergence,  405 

Division  of  straight  line,  8 

e,  the  number,  53 
Eccentricity  of  conic,  74,  78,  83 
Element  of  definite  integral.  2H0 
Ellipse,  74,  108 

area  of,  262 
Ellipsoid,  307 

volume  of,  270 
Epicycloid,  111 
Epi  trochoid,  114 
Equations,  differential,  438 

empirical,  89 

parametric,  106 

roots  of,  34 

Factor,  integrating,  448 
Focus  of  coiii,-,  74,  78,  80,  83 
Force,  179 

Forms,  indeterminate,  423.   12"> 
Fourier's  series,  427 
Fractions,  partial,  247 
Function,  9,  300 

Bessel's,  471 

complementary,  456,  458,  461 

composite,  357 

implicit,  163,  300 

notation,  12 

periodic,  427 

transcendental,  49 

Graph,  10,  20,  49,  301 

Helix,  324 
Hyperbola,  77 
Hyperboloid,  306,  307 
Hypocycloid,  112 

four-cusped,  107 
Hypotrochoid,  114 


Increment,  135,  339 
Infinitesimal,  261 
Infinity,  25 

Inflection,  point  of,  167 
Integral,  146,  222 

constant  of,  147,  222 

definite,  147,  260 

line,  353 

multiple.  369 

particular,  456,  458,  461 

triple,  385 
Integrand,  222,  264 
Integration,  222 

approximate,  41H 

collected  formulas  of,  236 

by  partial  fraction.-.  2  17 

by  paiis.  21:; 

of  a  polynomial.  1  Pi 

by  reduction  formulas,  252 

by  substitution,  238 
Intercepts,  21 
Intersections,  2!» 

Involute  of  circle,  1 12 

Legendre's  coefficients,  169 
Legendre's  equation,  168 
Lemniscate,  120,  L25 
Length  of  a  curve,  272.  325 
Limacon,  123 
Limit,  130 

of  ratio  of  arc  to  chord,  172 

Of  ^\    1,2 
h 

of  (i+  ft)*,  iog 

theorems  on,  132 
Limits  of  definite  integral,  148,  264 
Line,  straight,  57,  121,  317 
Line  integral,  353 
Locus,  21).  <;'.> 
Logarithm,  Napierian,  54 

Maclaurin's  series,  412 
Maxima  and  minima,  168,  348 
Mean,  theorem  of,  422 
Moment  of  inertia,  377,  390 


516 


INDEX 


Motion,  in  a  curve,  180 
rectilinear,  177 
simple  harmonic,  203 

Normal,  to  curve,  104,  326 

to  plane,  310 

to  surface,  347 
Number  scale,  3 

Operator,  455 

Order  of  differential  equation,  441 

Ordinate,  5 

Origin,  3,  118 

Parabola,  80 

segment  of,  83 
Paraboloid,  305,  308 
Parameter,  106 
Parts,  integration  by,  243 
Plane,  305,  316 
Point,  of  division,  8 

turning,  139 
Pole,  118 
Polynomial,  derivative  of,  137 

integral  of,  146 
Pressure,  275 

center  of,  277 
Prismoid,  421 
Prismoidal  formula,  419 
Projectile,  path  of,  181 
Projection,  2,  310,  323 

Radius  of  curvature,  208 

Radius  vector,  118 

Rate  of  change,  175 

Ratio  test,  407 

Reduction  formulas,  252 

Region  of  convergence,  411 

Remainder  in  Taylor's  series,  416 

Revolution,  surface  of,  270,  274,  309 

Rolle's  theorem,  416 

Roots  of  an  equation,  34 

Rose  of  three  leaves,  119 


Segment,  parabolic,  83 
Series,  405 

Fourier's,  427 

geometric,  405 

harmonic,  406 

Maclaurin's,  412 

power,  410 

Taylor's,  412 
Simpson's  rule,  421 
Slope,  6,  134 
Space  geometry,  300 
Sphere,  310,  313 
Spirals,  120 
Strophoid,  86 

Substitution,  integration  by,  238 
Surfaces,  304 

of  revolution,  270,  274,  309 
Symmetry,  23 

Tangent,  to  plane  curve,  30,  140,  164 

to  space  curve,  326 

to  surface,  345 
Taylors  series,  412 
Tractrix,  439 
Transformation  of  coordinates,  39, 

124,  386,  387,  388 
Trapezoidal  rule,  421 
Trochoid,  110 

Value,  absolute,  409 
infinite,  25 
mean,  265 

Variable,  9 

Vector,  radius,  118 

Velocity,  177,  180 

Vertex  of  conic,  75,  79,  81 

Volume,  of  any  solid,  389 
element  of,  385,  386,  387 
of  solid  with  parallel  bases,  268 
of  solid  of  revolution,  270 

Witch,  84 

Work,  275,  353,  356 


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